Direction Cosines And Directions Ratios

Let us assume the angles that made with the positive direction of x, y, and z-axes be ,, .

Then we get,

⇒

⇒

⇒

We know that if a line makes angles of with the positive x, y, and z-axes then the direction cosines of that line is the cosine of that angles made by that line with the axes.

Let us assume that l, m, n are the direction cosines of the line. Then,

⇒

⇒

⇒

We substitute the values of in the above equations for the values of l, m, n.

⇒

⇒

⇒

⇒

⇒

⇒

∴ The direction cosines of the given line is .

Let us assume the direction ratios of the line be r₁, r₂, r₃.

Then:

⇒ r₁ = 2

⇒ r₂ = –1

⇒ r₃ = –2

Let us assume the direction cosines for the line be l, m, n

We know that for a line of direction ratios r₁, r₂, r₃ and having direction cosines l, m, n has the following property.

⇒

⇒

⇒

Let us substitute the values of r₁, r₂, r₃ to find the values of .

⇒

⇒

⇒

⇒

⇒

⇒

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⇒

⇒

⇒

⇒

⇒

∴ The direction cosines for the given line is .

Let us assume the given two points of line be X(–2,4,–5) and Y(1,2,3).

Let us also assume the direction ratios for the given line be (r₁, r₂, r₃).

We know that direction ratios for a line passing through points (x₁, y₁, z₁) and (x₂, y₂, z₂) is (x₂–x₁, y₂–y₁, z₂–z₁).

So, using this property the direction ratios for the given line is, ⇒ (r₁, r₂, r₃) = (1–(–2), 2–4, 3–(–5))

⇒ (r₁, r₂, r₃) = (1+2, 2–4, 3+5)

⇒ (r₁, r₂, r₃) = (3, –2, 8)

Let us assume be the direction cosines of the given line.

We know that for a line of direction ratios r₁, r₂, r₃ and having direction cosines has the following property.

⇒

⇒

⇒

Let us substitute the values of r₁, r₂, r₃ to find the values of l, m, n.

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

∴ The Direction Cosines for the given line is .

Given points are:

⇒ A = (2,3,–4)

⇒ B = (1,–2,3)

⇒ C = (3,8,–11)

We know that for points D, E, F to be collinear the direction ratios of any two lines from DE, DF, EF are to be proportional;

We know that direction ratios for a line passing through points (x₁, y₁, z₁) and (x₂, y₂, z₂) is (x₂–x₁, y₂–y₁, z₂–z₁).

Let us assume direction ratios for AB is (r₁, r₂, r₃) and BC is (r₄, r₅, r₆).

The proportional condition can be stated as .

Let us find the direction ratios of AB

⇒ (r₁, r₂, r₃) = (1–2, –2–3, 3–(–4))

⇒ (r₁, r₂, r₃) = (1–2, –2–3, 3+4)

⇒ (r₁, r₂, r₃) = (–1, –5, 7)

Let us find the direction ratios of BC

⇒ (r₄, r₅, r₆) = (3–1, 8–(–2), –11–3)

⇒ (r₄, r₅, r₆) = (3–1, 8+2, –11–3)

⇒ (r₄, r₅, r₆) = (2, 10, –14)

Now

⇒ ……(1)

⇒

⇒ ……(2)

⇒

⇒ ……(3)

From (1),(2),(3) we get,

⇒

So, from the above relational we can say that points A, B , C are collinear.

Let us write the given points as:

⇒ A = (3,5,–4)

⇒ B = (–1,1,2)

⇒ C = (–5,–5,–2)

Let us assume the direction ratios of sides AB be (r₁,r₂,r₃), BC be (r₄,r₅,r₆) and CA be (r₇,r₈,r₉)

We know that direction ratios for a line passing through points (x₁, y₁, z₁) and (x₂, y₂, z₂) is (x₂–x₁, y₂–y₁, z₂–z₁).

Let us find the direction ratios for the side AB

⇒ (r₁,r₂,r₃) = (–1–3, 1–5, 2–(–4))

⇒ (r₁,r₂,r₃) = (–1–3, 1–5, 2+4)

⇒ (r₁,r₂,r₃) = (–4,–4,6)

Let us find the direction ratios for the side BC

⇒ (r₄,r₅,r₆) = (–5–(–1), –5–1, –2–2)

⇒ (r₄,r₅,r₆) = (–5+1, –5–1, –2–2)

⇒ (r₄,r₅,r₆) = (–4,–6,–4)

Let us find the direction ratios for the side CA

⇒ (r₇,r₈,r₉) = (3–(–5), 5–(–5), –4–(–2))

⇒ (r₇,r₈,r₉) = (3+5, 5+5, –4+2)

⇒ (r₇,r₈,r₉) = (8,10,–2)

Let us assume be the direction cosines of line AB, be the direction cosines of line BC and be the direction cosines of line CA.

We know that for a line of direction ratios r₁, r₂, r₃ and having direction cosines has the following property.

⇒

⇒

⇒

Let us follow the above property and find the direction cosines of each side.

Now, let’s find the direction cosines of side AB,

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⇒

⇒

⇒

⇒

⇒

The direction cosines for the side AB is .

Let’s find the directional cosines for the side BC,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

The direction cosines for the sides BC is .

Let’s find the direction cosines for the side CA,

⇒

⇒

⇒

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⇒

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⇒

⇒

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The direction cosines for the sides CA is .

Let us assume the direction ratios of vectors be (r₁,r₂,r₃) and (r₄,r₅,r₆).

Then,

⇒ (r₁,r₂,r₃) = (1,–2,1)

⇒ (r₄,r₅,r₆) = (4,3,2)

We know that the angle between the vectors with direction ratios proportional to (a₁,b₁,c₁) and (a₂,b₂,c₂) is given by:

⇒

Using the above formula we calculate the angle between the vectors.

Let be the angle between the two vectors given in the problem.

⇒

⇒

⇒

⇒

⇒

∴ The angle between two given vectors is .

Let us assume the direction ratios of vectors be (r₁,r₂,r₃) and (r₄,r₅,r₆).

Then,

⇒ (r₁,r₂,r₃) = (2,3,–6)

⇒ (r₄,r₅,r₆) = (3,–4,5)

We know that the angle between the vectors with direction ratios proportional to (a₁,b₁,c₁) and (a₂,b₂,c₂) is given by:

⇒

Using the above formula we calculate the angle between the vectors.

Let be the angle between the two vectors given in the problem.

⇒

⇒

⇒

⇒

⇒

⇒

∴ The angle between two given vectors is .

Given that the direction ratios of the lines are proportional to 2:3:6 and 1:2:2.

Let us denote the lines in the form of vectors as A and B.

Let’s write the vectors:

⇒ A = 2i + 3j + 6k

⇒ B = 1i + 2j + 2k

We know that the angle between the vectors a₁i + b₁j + c₁k and a₂i + b₂j + c₂k is given by:

⇒

Let’s assume the angle between the vectors A and B be ,

Using the given formula we find the value of .

⇒

⇒

⇒

⇒

⇒

The acute angle between the two vectors is given by .

Let us indicate given points with A, B and C.

⇒ A = (2,3,4)

⇒ B = (–1,–2,1)

⇒ C = (5,8,7)

We know that for points D, E, F to be collinear the direction ratios of any two lines from DE, DF, EF are to be proportional;

We know that direction ratios for a line passing through points (x₁, y₁, z₁) and (x₂, y₂, z₂) is (x₂–x₁, y₂–y₁, z₂–z₁).

Let us assume direction ratios for AB is (r₁, r₂, r₃) and BC is (r₄, r₅, r₆).

The proportional condition can be stated as .

Let us find the direction ratios of AB

⇒ (r₁, r₂, r₃) = (–1–2, –2–3, 1–4)

⇒ (r₁, r₂, r₃) = (–3,–5,–3)

Let us find the direction ratios of BC

⇒ (r₄, r₅, r₆) = (5–(–1), 8–(–2), 7–1)

⇒ (r₄, r₅, r₆) = (5+1, 8+2, 7–1)

⇒ (r₄, r₅, r₆) = (6, 10, 6)

Now

⇒

⇒ ……(1)

⇒

⇒ ……(2)

⇒

⇒ ……(3)

From (1),(2),(3) we get,

⇒

So, from the above relational we can say that points (2,3,4), (–1,–2,1) , (5,8,7) are collinear.

Let us denote the points as follows:

⇒ A = (4,7,8)

⇒ B = (2,3,4)

⇒ C = (–1,–2,1)

⇒ D = (1,2,5)

If two lines are said to be parallel the directional ratios of two lines need to be proportional.

Let us assume the direction ratios for line AB be (r₁,r₂,r₃) and CD be (r₄,r₅,r₆)

We know that direction ratios for a line passing through points (x₁, y₁, z₁) and (x₂, y₂, z₂) is (x₂–x₁, y₂–y₁, z₂–z₁).

Let’s find the direction ratios for the line AB

⇒ (r₁,r₂,r₃) = (2–4, 3–7, 4–8)

⇒ (r₁,r₂,r₃) = (–2,–4,–4)

Let’s find the direction ratios for the line CD

⇒ (r₄,r₅,r₆) = (1–(–1), 2–(–2), 5–1)

⇒ (r₄,r₅,r₆) = (1+1, 2+2, 5–1)

⇒ (r₄,r₅,r₆) = (2,4,4)

The proportional condition can be stated as .

Let check whether the directional ratios are proportional or not,

⇒

⇒ ……(1)

⇒

⇒ ……(2)

⇒

⇒ ……(3)

From (1),(2),(3) we can say that the direction ratios of the lines are proportional. So, the lines are parallel to each other.

Let us denote the points as follows:

⇒ A = (1,–1,2)

⇒ B = (3,4,–2)

⇒ C = (0,3,2)

⇒ D = (3,5,6)

If two lines of direction ratios (a₁,b₁,c₁) and (a₂,b₂,c₂) are said to be perpendicular to each other. Then the following condition is need to be satisfied:

⇒ a₁.a₂+b₁.b₂+c₁.c₂=0 ……(1)

Let us assume the direction ratios for line AB be (r₁,r₂,r₃) and CD be (r₄,r₅,r₆)

We know that direction ratios for a line passing through points (x₁, y₁, z₁) and (x₂, y₂, z₂) is (x₂–x₁, y₂–y₁, z₂–z₁).

Let’s find the direction ratios for the line AB

⇒ (r₁,r₂,r₃) = (3–1, 4–(–1), –2–2)

⇒ (r₁,r₂,r₃) = (3–1, 4+1, –2–2)

⇒ (r₁,r₂,r₃) = (2,5,–4)

Let’s find the direction ratios for the line CD

⇒ (r₄,r₅,r₆) = (3–0, 5–3, 6–2)

⇒ (r₄,r₅,r₆) = (3,2,4)

Let us check whether the lines are perpendicular or not using (1)

⇒ r₁.r₄+r₂.r₅+r₃.r₆ = (2×3)+(5×2)+(–4×4)

⇒ r₁.r₄+r₂.r₅+r₃.r₆ = 6+10–16

⇒ r₁.r₄+r₂.r₅+r_3.r₆ = 0

Since the condition is clearly satisfied, we can say that the given lines are perpendicular to each other.

Let us denote the points as follows:

⇒ O = (0,0,0)

⇒ A = (2,1,1)

⇒ B = (3,5,–1)

⇒ C = (4,3,–1)

If two lines of direction ratios (a₁,b₁,c₁) and (a₂,b₂,c₂) are said to be perpendicular to each other. Then the following condition is need to be satisfied:

⇒ a₁.a₂+b₁.b₂+c₁.c₂=0 ……(1)

Let us assume the direction ratios for line OA be (r₁,r₂,r₃) and BC be (r₄,r₅,r₆)

We know that direction ratios for a line passing through points (x₁, y₁, z₁) and (x₂, y₂, z₂) is (x₂–x₁, y₂–y₁, z₂–z₁).

Let’s find the direction ratios for the line OA

⇒ (r₁,r₂,r₃) = (2–0, 1–0, 1–0)

⇒ (r₁,r₂,r₃) = (2,1,1)

Let’s find the direction ratios for the line BC

⇒ (r₄,r₅,r₆) = (4–3, 3–5, –1–(–1))

⇒ (r₄,r₅,r₆) = (4–3, 3–5, –1+1)

⇒ (r₄,r₅,r₆) = (1,–2,0)

Let us check whether the lines are perpendicular or not using (1)

⇒ r₁.r₄+r₂.r₅+r₃.r₆ = (2×1)+(1×–2)+(1×0)

⇒ r₁.r₄+r₂.r₅+r₃.r₆ = 2–2+0

⇒ r₁.r₄+r₂.r₅+r_3.r₆ = 0

Since the condition is clearly satisfied, we can say that the given lines are perpendicular to each other.

Let us assume the direction ratios of vectors be (r₁,r₂,r₃) and (r₄,r₅,r₆).

Then,

⇒ (r₁,r₂,r₃) = (a,b,c)

⇒ (r₄,r₅,r₆) = (b–c, c–a, a–b)

We know that the angle between the lines with direction ratios proportional to (a₁,b₁,c₁) and (a₂,b₂,c₂) is given by:

⇒

Using the above formula we calculate the angle between the lines.

Let be the angle between the two lines given in the problem.

⇒

⇒

⇒

⇒

⇒

∴ The angle between two given vectors is .

Given points are:

⇒ A = (1,2,3)

⇒ B = (4,5,7)

⇒ C = (–4,3,–6)

⇒ D = (2,9,2)

Let us assume the direction ratios for line AB be (r₁,r₂,r₃) and CD be (r₄,r₅,r₆)

We know that direction ratios for a line passing through points (x₁, y₁, z₁) and (x₂, y₂, z₂) is (x₂–x₁, y₂–y₁, z₂–z₁).

Let’s find the direction ratios for the line AB

⇒ (r₁,r₂,r₃) = (4–1, 5–2, 7–3)

⇒ (r₁,r₂,r₃) = (3,3,4)

Let’s find the direction ratios for the line CD

⇒ (r₄,r₅,r₆) = (2–(–4), 9–3, 2–(–6))

⇒ (r₄,r₅,r₆) = (2+4, 9–3, 2+6)

⇒ (r₄,r₅,r₆) = (6,6,8)

We know that the angle between the vectors with direction ratios proportional to (a₁,b₁,c₁) and (a₂,b₂,c₂) is given by:

⇒

Using the above formula we calculate the angle between the vectors.

Let be the angle between the two vectors given in the problem.

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

∴ The angle between the given two vectors is 0⁰.

Given relations are:

⇒ 2lm+ 2ln– mn =0 ……(1)

⇒ l+ m+ n =0

⇒ l = (–m– n) ……(2)

Substituting (2) in (1) we get,

⇒ 2(–m–n)m + 2(–m–n)n – mn =0

⇒ 2(–m²–mn) + 2(–mn–n²) – mn =0

⇒ –2m² –2mn –2mn –2n² –mn =0

⇒ –2m²–5mn–2n² =0

⇒ 2m²+5mn+2n²=0

⇒ 2m²+4mn+mn+2n²=0

⇒ 2m(m+2n)+n(m+2n)=0

⇒ (2m+n)(m+2n)=0

⇒ 2m+n=0 or m+2n=0

⇒ 2m=–n or m=–2n

⇒ ……(3)

Substituting the values of (3) in eq(2), we get

For 1^st line:

⇒

⇒

⇒

The direction ratios for the first line is .

Let us assume l₁,m₁,n₁ be the direction cosines of 1^st line.

We know that for a line of direction ratios r₁, r₂, r₃ and having direction cosines has the following property.

⇒

⇒

⇒

Using the above formulas we get,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

The Direction cosines for the 1^st line is

For 2^nd line:

⇒ l=–(–2n)–n

⇒ l=2n–n

⇒ l=n

The direction ratios for the second line is .

Let us assume l₂,m₂,n₂ be the direction cosines of 1^st line.

We know that for a line of direction ratios r₁, r₂, r₃ and having direction cosines has the following property.

⇒

⇒

⇒

Using the above formulas we get,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

The Direction Cosines for the 2^nd line is .

Given relations are:

⇒ l²+m²–n²=0 ……(1)

⇒ l+m+n=0

⇒ l=–m–n……(2)

Substituting (2) in (1) we get,

⇒ (–m–n)²+m²–n²=0

⇒ m²+n²+2mn+m²–n²=0

⇒ 2m²+2mn=0

⇒ 2m(m+n)=0

⇒ 2m=0 or m+n=0

⇒ m=0 or m=–n ……(3)

Substituting value of m from(3) in (2)

For the 1^st line:

⇒ l=–0–n

⇒ l=–n

The Direction Ratios for the first line is (–n,0,n)

For the 2^nd line:

⇒ l=–(–n)–n

⇒ l=n–n

⇒ l=0

The Direction Ratios for the second line is (0,–n,n)

We know that the angle between the lines with direction ratios proportional to (a₁,b₁,c₁) and (a₂,b₂,c₂) is given by:

⇒

Using the above formula we calculate the angle between the lines.

Let be the angle between the two lines given in the problem.

⇒

⇒

⇒

⇒

⇒

∴ The angle between given two lines is .

Given relations are:

⇒ mn+nl+lm=0 ……(1)

⇒ 2l–m+2n=0

⇒ m=2l+2n ……(2)

Substituting (2) in (1) we get,

⇒ (2l+2n)n+nl+l(2l+2n)=0

⇒ 2ln+2n²+nl+2l²+2ln=0

⇒ 2n²+5ln+2l²=0

⇒ 2n²+4ln+ln+2l²=0

⇒ 2n(n+2l)+l(n+2l)=0

⇒ (2n+l)(n+2l)=0

⇒ 2n+l=0 or n+2l=0

⇒ l=–2n or 2l=–n ……(3)

Substituting the values of(3) in (2) we get,

For the 1^st line:

⇒ m = 2(–2n)+2n

⇒ m=–4n+2n

⇒ m=–2n

The direction ratios for the 1^st line is (–2n,–2n,n)

For the 2^nd line:

⇒ m=–n+2n

⇒ m=n

The direction ratios for the 2^nd line is

We know that the angle between the lines with direction ratios proportional to (a₁,b₁,c₁) and (a₂,b₂,c₂) is given by:

⇒

Using the above formula we calculate the angle between the lines.

Let be the angle between the two lines given in the problem.

⇒

⇒

⇒

⇒

⇒

∴ the angle between two lines is .

Given relations are:

⇒ 3lm–4ln+mn=0 ……(1)

⇒ l+2m+3n=0

⇒ l=–2m–3n ……(2)

Substituting (2) in (1) we get,

⇒ 3(–2m–3n)m –4(–2m–3n)n +mn =0

⇒ 3(–2m²–3mn) –4(–2mn–3n²) +mn=0

⇒ –6m²–9mn+8mn+12n²+mn=0

⇒ 12n²–6m²=0

⇒ m²–2n²=0

⇒

⇒

⇒ ……(3)

Substituting the values of (3) in (2) we get,

For the 1^st line:

⇒

⇒

The Direction Ratios for the 1^st line is .

For the 2^nd line:

⇒

⇒

The Direction Ratios for the 2^nd line is .

We know that the angle between the lines with direction ratios proportional to (a₁,b₁,c₁) and (a₂,b₂,c₂) is given by:

⇒

Using the above formula we calculate the angle between the lines.

Let be the angle between the two lines given in the problem.

⇒

⇒

⇒

⇒

⇒

∴ The angle between two lines is .

Given relations are:

⇒ mn+ln+lm=0 ……(1)

⇒ 2l+2m–n=0

⇒ n=2l+2m ……(2)

Substituting (2) in (1) we get,

⇒ m(2l+2m)+l(2l+2m)+lm=0

⇒ 2lm+2m²+2l²+2lm+lm=0

⇒ 2m²+5lm+2l²=0

⇒ 2m²+4lm+lm+2l²=0

⇒ 2m(m+2l)+l(m+2l)=0

⇒ (2m+l)(m+2l)=0

⇒ 2m+l=0 or m+2l=0

⇒ 2m=–l or 2l=–m ……(3)

Substituting the values of (3) in (2), we get

For the 1^st line:

⇒ n=2l–l

⇒ n=l

The Direction Ratios for the first line is

For the 2^nd line:

⇒ n=–m+2m

⇒ n=m

The Direction Ratios for the second line is

We know that the angle between the lines with direction ratios proportional to (a₁,b₁,c₁) and (a₂,b₂,c₂) is given by:

⇒

Using the above formula we calculate the angle between the lines.

Let be the angle between the two lines given in the problem.

⇒

⇒

⇒

⇒

∴ the angle between two lines is .

The direction cosines of a directed line can be defined as cosine values of the angles made by the directed line with the x-axis, y-axis and z-axis respectively.

Explanation:

Consider a directed line , in the three dimensional space.

If α, β and γ be the angles made by the directed line with the x-axis, y-axis and z-axis respectively.

In the above figure, the direction cosines of line OA are:

Cos α = cosine of the angle between x-axis (OX) and the directed line.

Cos β = cosine of the angle between y-axis (OY) and the directed line.

Cos γ = cosine of the angle between z-axis (OZ) and the directed line.

As per the definition of direction cosines, the cosine values of the angles formed by the directed line with the x-axis, y-axis and z-axis.

Here we consider the directed line to be the x-axis.

So from the below figure, we can say,

α = the angle formed by the x-axis with x-axis = 0°

β = the angle formed by the x-axis with y-axis = 90°

γ = the angle formed by the x-axis with y-axis = 90°

Therefore,

cos α = cos 0° = 1

cos β = cos 90° = 0

cos γ = cos 90° = 0

Hence the direction cosines of x-axis are 1, 0, 0.

As per the definition of direction cosines, the cosine values of the angles formed by the directed line with the x-axis, y-axis and z-axis.

Here we consider the directed line to be the y-axis.

So from the below figure, we can say,

α = the angle formed by the y-axis with x-axis = 90°

β = the angle formed by the y-axis with y-axis = 0°

γ = the angle formed by the y-axis with y-axis = 90°

Therefore,

cos α = cos 90° = 0

cos β = cos 0° = 1

cos γ = cos 90° = 0

Hence the direction cosines of y-axis are 0, 1, 0.

As per the definition of direction cosines, the cosine values of the angles formed by the directed line with the x-axis, y-axis and z-axis.

Here we consider the directed line to be the z-axis.

So from the below figure, we can say,

α = the angle formed by the z-axis with x-axis = 90°

β = the angle formed by the z-axis with y-axis = 90°

γ = the angle formed by the x-axis with y-axis = 0°

Therefore,

cos α = cos 90° = 0

cos β = cos 90° = 0

cos γ = cos 0° = 1

Hence the direction cosines of y-axis are 0, 0, 1.

From the given information, A is a point with co-ordinates (3,-2, 3).

If you consider the projection of A(3,-2,3) on the XY-plane is H(3,-2,0) where the z-coordinate will not exist on XY-plane.

Similarly projection of A(3,-2,3) on the YZ-plane is T(0,-2,3) where the x-coordinate will not exist on YZ-plane.

The projection of A(3,-2,3) on the XZ-plane is T(3,0,3) where the x-coordinate will not exist on XZ-plane.

Now, the distance between A and XY-plane = Distance between points A&H

Distance between two points is given by

Using this formula,

Distance of point A from XY

= √ 3²

= 3

Distance of point A from YZ

= √ 3²

= 3

Distance of point A from XZ

= √ 2²

= 2

From the given information, A is a point with co-ordinates (3, -5, 12).

From the figure, we can say that the projection of point A on x-axis will be point H(3,0,0) as the y-coordinate and z-coordinate will be zeros.

Distance between two points is given by

Using this formula,

Distance of point A from x-axis (point H)

=√ 169

=13

Given the points P(-2,5,9) and Q(3,-2,4)

Let the plane YZ-plane divide line segment PQ at point G(0,y,z) in the ratio m:n.

The coordinates of the point G which divides the line joining points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) in the ratio m:n is given by

Here, we have m:n

x₁ = -2 y₁ = 5 z₁ = 9

x₂ = 3 y₂ = -2 z₂ = 4

By using the above formula, we get,

Now, this is the same point as G(0,y,z),

As the x-coordinate is zero,

[Cross Multiplying]

3m – 2n = 0 × (m + n)

3m – 2n = 0

3m = 2n

Therefore, the ratio in which the plane-YZ divides the line joining A & B is 2:3

Given that, the line makes angles

• 60° with the x-axis.

• 60° with the y-axis.

Let the angle made by the line with z-axis be α.

Now, as per the relation between direction cosines of a line, l² + m²+n² = 1 where l,m,n are the direction cosines of a line from x-axis, y-axis and z-axis respectively.

From the problem,

l = cos 60°

m = cos 60°

n = cos α

By using the formula,

l² + m²+n² = 1

[As cos 60° value is ]

[As cos 45° ]

α = 45°

Therefore, the angle made by the line with z-axis is 45°

Given, the line makes the angles α, β and γ respectively with x-axis, y-axis and z-axis.

As per the relation between direction cosines of a line, l² + m²+n² = 1 where l,m,n are the direction cosines of a line from x-axis, y-axis and z-axis respectively.

So, we can say that,

cos²α + cos²β + cos²γ = 1 ------ (1)

Now, we should find the value for

cos2α + cos2β + cos2γ

cos2α can be written as 2cos²α -1,

cos2α + cos2β + cos2γ = (2cos²α -1) + (2cos²β -1) + (2cos²γ -1)

= 2 (cos²α+ cos²β + cos²γ) – 3

= 2(1) – 3

[From Equation (1)]

= -1

Therefore,

cos2α + cos2β + cos2γ = -1

Given,

The line segment is formed by P and Q points where

Point P = (a,b,c)

Point Q = (-a,-c,-b)

From the figure, we can clearly see that, the line segment joining points P and Q is meeting the plane XY at point G.

Let Point G be (x,y,0) as the z-coordinate on xy plane does not exist.

Also let point G divides the line segment joining P and Q in the ratio m:n.

The coordinates of the point G which divides the line joining points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) in the ratio m:n is given by

Here, we have m:n

x₁ = a y₁ = b z₁ = c

x₂ = -a y₂ = -c z₂ = -b

By using the above formula, we get,

Now, this is the same point as G(x,y,0),

As the x-coordinate is zero,

[Cross Multiplying]

-bm + cn = 0 × (m + n)

-bm + cn = 0

-bm = -cn

Therefore, the ratio in which the plane-XY divides the line joining P & Q is c:b

Given, the direction ratios of the line are proportional to (0, 1,-1)

Therefore, consider the direction ratios of the give line can be

a = 0 × k, b = 1 × k, c = (-1) × k

[where k is some proportionality constant]

Now the direction ratios of the line are

a = 0, b = k, c = -k

As we know the direction cosine of z-axis can be given by

cos γ = n where γ is the angle made by the line with the z-axis.

By using the above formula:

[As cosine function is negative, the angle become 135° instead of 45° ]

The inclination of the line with z-axis is

Given,

• Direction Ratios of Line1 are proportional to (1,-2,1)

• Direction Ratios of Line2 are proportional to (4,3,2)

So we can say that,

Direction ratios of line1

a₁ = 1 × k , b₁ = (-2) × k and c₁ = 1 × k

a₁ = k , b₁ = -2k and c₁ = k

Direction ratios of line2

a₂ = 4 × p , b₂ = 3 × p and c₂ = 2 × p

a₂ = 4p , b₂ = 3p and c₂ = 2p

Now, the angle between the lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ is given by

By using this formula,

cos θ = 0

θ = 90°

The angle between the lines is 90°.

Given point P(x,y,z)

From the figure, we can say that Point E (x,y,0) is the projection of Point P on the XY-plane ( the z-coordinate remains zero on XY-plane).

Distance between two points is given by

Here the distance between Point P & E will give the distance of the point P from the XY-plane.

Here a₁ = x, b₁ =y , c₁ =z

a₂ = x, b₂ = y, c₂ = 0

Distance from P to E =

= √ (-z)²

= √ (z)²

= z

Therefore, the distance between the XY plane and point P is z units.

Given, point P (x,y,z)

From the figure, we can clearly see the projection of point P on the XOZ plane.

The projection of P on the x-axis will be (x,0,0)

The projection of P on the z-axis will be (0,0,z)

By this we can say that, if we are considering the projection of P on the XOZ plane, the coordinates of Y-axis will be zero,

Hence the projection of point P(x,y,z) on the XOZ plane will be point E(x,o,z).

Given Point P is (2,-3,5)

From the figure, we can see that Point E is the projection of P (2,-3,5) on the Y-axis.

All the points on the y-axis are of the form (0,y,0).

Hence, the projection of point P on y-axis will be (0,-3,0).

Given,

The point is (2,3,4). Let this point be P.

From the figure, point E (2,0,0) is the projection of point P(2,3,4) on the x-axis.

The distance between the points P & E will give the distance of the point P from x-axis.

Distance between two points is given by

Here

a₁ =2 , b₁ =3 , c₁=4 and a₂ =2 , b₂ =0 , c₂ = 0

Distance between P and x-axis is

= √ 25

= 5

Therefore the distance between, the x-axis and the Point P (2,3,4) is 5 units.

Given, the direction ratios of the line are proportional to (2, -1,-2)

Therefore, consider the direction ratios of the give line can be

a = 2 × k, b = (-1) × k, c = (-2) × k

[where k is some proportionality constant]

Now the direction ratios of the line are

a = 2k, b = -k, c = -2k

As we know the direction cosine ae given by

cos α = l , cos β = m , cos γ = n

Where α, β and γ are the angles formed by the line with the three axes.

By using the above formula:

l = cos α =

Therefore cos α

m =cos β =

cos β

n = cos γ =

cos γ

Therefore, the direction cosines are

Given

The line is parallel to z- axis.

So the line would be perpendicular to both x-axis and y-axis.

Hence, the angles formed by the line with x-axis & y-axis are 90° and 90° respectively.

Also the angle formed by the line with z-axis is 0°.

The direction cosines of a line are given by, cos α , cos β , cos γ. Where α,β and γ are angles formed by the line with the x,y and z axes respectively.

Here

α = 90°,β = 90° and γ = 0°

α = cos 90°= 0

β = cos 90°= 0

γ = cos 0° = 1

Therefore the direction cosines of the line parallel to z-axis are (0,0,1).

Given the unit vector makes,

• an angle of with x-axis

• an angle of with y-axis

• an angle of θ with z-axis

• θ is acute angle

Let the unit vector be:

As given it is a unit vector,

Therefore = 1

As the angle between in and x-axis is , the scalar product of the vectors can be performed.

The scalar product of the two vectors is given by

[as both the vectors are of magnitude 1].

As the angle between in and y-axis is , the scalar product of the vectors can be performed.

Similarly the angle between in and y-axis is θ , the scalar product of the vectors can be performed.

The magnitude of a vector x+ y+ zis given by.

Now consider the magnitude of the vector

1

1

[Squaring on both sides]

1 =

cos²θ

cos²θ =

cos θ =

cosθ =

As given in the question θ is acute angle, so θ belongs to 1^st quadrant and is positive.

Therefore

Given,

The point is (a,b,c). Let this point be P.

From the figure, point E (a,0,0) is the projection of point P(a,b,c) on the x-axis.

The distance between the points P & E will give the distance of the point P from x-axis.

Distance between two points is given by

Here

a₁ =a , b₁ =b , c₁=c and a₂ =a , b₂ =0 , c₂ = 0

Distance between P and x-axis is

Therefore the distance between, the x-axis and the Point P (a,b,c) is units.

Given a line makes,

• an angle of 90° with x-axis

• an angle of 60° with y-axis

So, let the angle made by the line with z-axis is θ

Now, as per the relation between direction cosines of a line, l² + m²+n² = 1 where l,m,n are the direction cosines of a line from x-axis, y-axis and z-axis respectively.

From the problem,

l = cos 90° = 0

m = cos 60°

n = cos θ

By using the formula,

l² + m²+n² = 1

As the angle made by the line with positive z-axis, so the cosine angle is positive.

Therefore, cos θ =

Hence θ = 30°.