Direction Cosines: Definition, Formula and Sample Questions

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Direction cosines as a concept are found in Cartesian geometry (on 3-dimensional Cartesian plane). The term Direction Cosines refers to the angles made with positive side of the axes by a given Position Vector with all the three axes viz. x, y, z which are known as Direction Angles α, β, γ respectively. Now, cosine of these angles i.e., cosα, cosβ, cosγ is known as Direction Cosines.

Table of Content

Position Vectors
Direction cosines
How to find direction cosines?
Direction ratios
Things to Remember
Sample Questions

Key Terms: Direction Cosines, Vectors, Magnitude, Angles, Cartesian Plane, axes, Cartesian Plane, Position Vector, Straight-line vector

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Position Vectors

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It is a straight-line vector whose one end is fixed to a body and the other end remains moving. In context of direction cosines, once we take a point on Cartesian Plane (say, Q), it is connected to the origin point of the axes and it forms a line OS, this is the position vector called OQ and this position vector will be used to make positive angles with the axes of cartesian plane which will be used to determine direction cosines.

OQ and OP are position Vectors.

Direction Cosines

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To understand Direction Cosines, consider the Position Vector OP (or r) which is made from placing a point P (x, y, z) in the 3D plane and connecting it to the origin point O. This position vector r will make angles α, β, γ with the positive side of the axes x, y, z respectively. These angles are known as Direction Angles. The cosine of these Direction Angles will be written as cosα, cosβ, cosγ and these are known as Direction Cosines of the vector r. The Cosα, Cosβ, Cosγ are also denoted as l, m, n respectively.

So, l = cosα m = cosβ n=cosγ

Read More: Displacement vector

How to determine Direction Cosines?

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To determine Direction Cosines, there is already the position vector present from when we identified Direction Cosines. Let’s call that position vector OP (or r). And the Direction Angles from the previous discussion were α, β, γ. Then the Direction Cosines of these angles would be cosα (or l), cosβ (or m), cosγ (or n) respectively.

Now, Direction Cosine is the division of the corresponding co-ordinate of the vector by the vector length. In order to determine the value of direction cosines plot the following points: A (on x-axis), B (on y Axis), C (on z axis). Now you can see all these points when connected with the point P form three separate right-angled triangles viz. ?OAP, ?OBP, and ?OCP where,

∠POA = α, ∠POB = β, ∠POC = γ

And coordinates of the vector (OP) are x, y, z

So, the Vector Length (of OP) = √x ²+ y² + z²

Now, cosα = l = xx²+y²+z²

Cosβ = m = yx²+y²+z²

Cosγ = n = zx²+y²+z²

This is the formula for finding direction cosines individually on each axis. In other words, Direction Cosine or cosine of any angle in a right-angled triangle will be: the length of adjacent side of the angle divided by the length of the hypotenuse.

For example, in the right-angled triangle ΔOAP

Here, the direction angle is ∠POA let’s call it α.

So, the direction cosine of this angle will be

Cosα = length of the adjacent side / length of the hypotenuse

So, cosα = OA/OP

Where OA is nothing but the value of x co-ordinate on point P and OP’s length is the magnitude of the vector OP.

Read More: Vector product of two vectors

Direction cosines in case a line does not pass-through origin

The line on which the position vector depends may not always pass through the origin. So, in order to determine the direction cosine with such a line, it is necessary to presume an imaginary line parallel to the given line where the 2nd line (or the imaginary line) passes through the origin. Now the direction angles made by this imaginary line will be the same as our given line so the formula given above to determine direction cosines can be used here to find the solution.

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Direction Ratios

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Direction Ratios of a line are any 3 numbers that are proportional to the direction cosines of a line.

Consider a vector OP=al+bJ+ck

Here, a, b, and c are Direction Ratios

Read More: Types of Vector

Things to Remember

If we reverse the direction of the position vector then the direction angles will also be different and consequently their direction cosines will also be different.
The direction cosines of any line parallel to the co-ordinate axes will the same as that the co-ordinate axis they are parallel to (because the direction cosines of two parallel lines are similar to one another) and the direction cosine of X, Y, and Z axes are 1,0,0, 0,1,0, 0,0,1 respectively.
If the direction of the given line is reversed then direction cosines will become cosπ-α, cosπ-β, cosπ-γ
The relationship between all here direction cosines of line can be formulated as

l² + m² + n² = 1

Read More: Co Planar Vectors

Sample Questions

Ques. Find the Direction Cosines of i+2J+3k . (1 mark)

Ans. Let r=l+2J+3k

So, the direction ratios are a = 1, b=2, c=3

Now magnitude of the vector r = √ 1² + 2² + 3² = √1 + 4 + 9 = √14

So, the direction cosines are ( ar, br, cr) or ( 1√14, 2√14, 3√14)

Ques. Find the direction cosines of the vector joining the points A (1, 2, –3) and B (–1, –2, 1), directed from A to B. (2 mark)

Ans. A (1, 2, -3)

B (-1, -2, 1)

AB = -1-1l+(-2-2)J+(1-(-3))k

AB = -2l-4J+4k

Direction ratios are, a = -2, b= -4, c= 4

Magnitude of the vector AB = √ (-2)² + (-4)² + 4²

| AB | = √ 4 + 16 + 16 = √36 = 6.

Direction Cosines are (aAB, bAB, cAB) or ( -26, -46, 46) = ( -13, -23, 23)

Ques. What are direction angles? (1 mark)

Ans. Direction angles are those angles which are made the given position vector with the co-ordinate axis in its positive direction. These angles will change if the direction of the given vector is reversed.

Ques. What is a position vector? (1 mark)

Ans. A position vector is a line with one end fixed to some point and the other end is free moving. It is used to describe position of a point in Cartesian plane.

Ques. What is a direction cosine? (1 mark)

Ans. A direction cosine is cos of a Direction Angle. In other words, we can say that it is the division of length of side adjacent to the direction angle and the length of hypotenuse in a right-angled triangle.

Ques. What is direction cosine in terms of sine? (1 mark)

Ans. The direction cosine in terms of sine can be expressed as

Sinθ= √ 1- cos²θ

Where, θ can be α, β, or γ i.e., any of the direction angles.

Ques. What is the direction cosine between two lines making angle θ? (1 mark)

Ans. Let there be two lines L₁ and L₂ with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ respectively.

If there are two lines placed so that the angle between them is θ then the formula for the direction cosine of such an angle will be:

Cosθ = (a₁a₂ + b₁b₂ + c₁c₂)

√a₁² + b₁² +c₁² √a₂² + b₂² +c₂ ²

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Direction Cosines: Definition, Formula and Sample Questions

Position Vectors

Direction Cosines

How to determine Direction Cosines?

Direction Ratios

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
Let A=\(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\),show that(aI+bA)ⁿ=aⁿI+na^n-1bA,where I is the identity matrix of order 2 and n∈N

2.
If A'= \(\begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 &1 \end{bmatrix}\)\(\begin{bmatrix} -1 & 2 & 1 \\ 1 &2 & 3\end{bmatrix}\) , then verify that
(i) \((A+B)'=A'+B' \)
(ii) \((A-B)'=A'-B'\)

3.
Evaluate \(\begin{vmatrix} cos\alpha cos\beta &cos\alpha sin\beta &-sin\alpha \\ -sin\beta&cos\beta &0 \\ sin\alpha cos\beta&sin\alpha\sin\beta &cos\alpha \end{vmatrix}\)

4.
Find the following integral: \(\int (ax^2+bx+c)dx\)

5.
Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)

6.
Let f: R→R be defined as f(x) = 3x. Choose the correct answer.

Similar Mathematics Concepts

Comments

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Direction Cosines: Definition, Formula and Sample Questions

Position Vectors

Direction Cosines

How to determine Direction Cosines?

Direction Ratios

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1. Let A=\(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\),show that(aI+bA)n=anI+nan-1bA,where I is the identity matrix of order 2 and n∈N

2. If A'= \(\begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 &1 \end{bmatrix}\)\(\begin{bmatrix} -1 & 2 & 1 \\ 1 &2 & 3\end{bmatrix}\) , then verify that (i) \((A+B)'=A'+B' \)(ii) \((A-B)'=A'-B'\)

3. Evaluate \(\begin{vmatrix} cos\alpha cos\beta &cos\alpha sin\beta &-sin\alpha \\ -sin\beta&cos\beta &0 \\ sin\alpha cos\beta&sin\alpha\sin\beta &cos\alpha \end{vmatrix}\)

4. Find the following integral: \(\int (ax^2+bx+c)dx\)

5. Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)

6. Let f: R→R be defined as f(x) = 3x. Choose the correct answer.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Let A=\(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\),show that(aI+bA)ⁿ=aⁿI+na^n-1bA,where I is the identity matrix of order 2 and n∈N

2.
If A'= \(\begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 &1 \end{bmatrix}\)\(\begin{bmatrix} -1 & 2 & 1 \\ 1 &2 & 3\end{bmatrix}\) , then verify that
(i) \((A+B)'=A'+B' \)
(ii) \((A-B)'=A'-B'\)

3.
Evaluate \(\begin{vmatrix} cos\alpha cos\beta &cos\alpha sin\beta &-sin\alpha \\ -sin\beta&cos\beta &0 \\ sin\alpha cos\beta&sin\alpha\sin\beta &cos\alpha \end{vmatrix}\)

4.
Find the following integral: \(\int (ax^2+bx+c)dx\)

5.
Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)

6.
Let f: R→R be defined as f(x) = 3x. Choose the correct answer.