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Vector Algebra - Direction cosines and direction ratios

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Direction Angles: When a vector r⃗\vec{r} is placed with its tail at the origin in a 3-dimensional Cartesian coordinate system, it makes specific angles α,β,\alpha, \beta, and γ\gamma with the positive directions of the x,y,x, y, and zz axes respectively. These are known as direction angles. Visually, α\alpha is the tilt toward the xx-axis, β\beta toward the yy-axis, and γ\gamma toward the zz-axis.

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Direction Cosines (DCs): The cosine values of the direction angles, denoted as l=cos⁡α,m=cos⁡β,l = \cos \alpha, m = \cos \beta, and n=cos⁡γn = \cos \gamma, are called the direction cosines of the vector. A critical property is that the sum of their squares is always equal to 1, represented by the identity l2+m2+n2=1l^2 + m^2 + n^2 = 1. This implies that the point (l,m,n)(l, m, n) always lies on a unit sphere centered at the origin.

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Direction Ratios (DRs): Any three numbers a,b,a, b, and cc that are proportional to the direction cosines l,m,l, m, and nn are called direction ratios of the vector. Unlike direction cosines, which are unique for a directed line, direction ratios are not unique; any scalar multiple (ka,kb,kc)(ka, kb, kc) also serves as a set of direction ratios for the same line. Visually, these ratios represent the components of a vector that is parallel to the original vector.

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Vector Components as DRs: For any vector given in the form r⃗=ai^+bj^+ck^\vec{r} = a\hat{i} + b\hat{j} + c\hat{k}, the coefficients a,b,a, b, and cc are the direction ratios of the vector. The magnitude of this vector is given by ∣r⃗∣=a2+b2+c2|\vec{r}| = \sqrt{a^2 + b^2 + c^2}.

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Relationship between DCs and DRs: If a,b,ca, b, c are the direction ratios of a vector, the direction cosines can be found by dividing each ratio by the magnitude of the vector. This process effectively 'normalizes' the direction ratios into a unit vector r^=li^+mj^+nk^\hat{r} = l\hat{i} + m\hat{j} + n\hat{k}.

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Direction Ratios of a Line Joining Two Points: For a line segment passing through two points P(x1,y1,z1)P(x_1, y_1, z_1) and Q(x2,y2,z2)Q(x_2, y_2, z_2), the direction ratios are calculated as the difference between their corresponding coordinates: (x2−x1),(y2−y1),(x_2 - x_1), (y_2 - y_1), and (z2−z1)(z_2 - z_1). This represents the displacement vector from PP to QQ.

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DCs of a Line: While a vector has a specific direction, a line extending in both directions has two sets of direction cosines: (l,m,n)(l, m, n) and (−l,−m,−n)(-l, -m, -n). This reflects the fact that the line can be viewed as being directed in either of two opposite ways along the same path.

📐Formulae

l=cos⁡α,m=cos⁡β,n=cos⁡γl = \cos \alpha, m = \cos \beta, n = \cos \gamma

l2+m2+n2=1l^2 + m^2 + n^2 = 1

l=±aa2+b2+c2,m=±ba2+b2+c2,n=±ca2+b2+c2l = \pm \frac{a}{\sqrt{a^2 + b^2 + c^2}}, m = \pm \frac{b}{\sqrt{a^2 + b^2 + c^2}}, n = \pm \frac{c}{\sqrt{a^2 + b^2 + c^2}}

r⃗=∣r⃗∣(li^+mj^+nk^)\vec{r} = |\vec{r}|(l\hat{i} + m\hat{j} + n\hat{k})

DRs of line segment PQ=(x2−x1,y2−y1,z2−z1)\text{DRs of line segment } PQ = (x_2 - x_1, y_2 - y_1, z_2 - z_1)

Unit vector r^=li^+mj^+nk^\text{Unit vector } \hat{r} = l\hat{i} + m\hat{j} + n\hat{k}

💡Examples

Problem 1:

Find the direction cosines of the vector a⃗=2i^+2j^−k^\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}.

Solution:

Step 1: Identify the direction ratios a,b,ca, b, c from the vector coefficients. Here, a=2,b=2,c=−1a = 2, b = 2, c = -1. Step 2: Calculate the magnitude of the vector: ∣a⃗∣=22+22+(−1)2=4+4+1=9=3|\vec{a}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 Step 3: Apply the formula for direction cosines l=a∣a⃗∣,m=b∣a⃗∣,n=c∣a⃗∣l = \frac{a}{|\vec{a}|}, m = \frac{b}{|\vec{a}|}, n = \frac{c}{|\vec{a}|}: l=23,m=23,n=−13l = \frac{2}{3}, m = \frac{2}{3}, n = \frac{-1}{3} Thus, the direction cosines are (23,23,−13)(\frac{2}{3}, \frac{2}{3}, -\frac{1}{3}).

Explanation:

To find the direction cosines, we first treat the scalar components of the vector as direction ratios. Dividing these ratios by the total magnitude of the vector scales them down to the values that represent the cosines of the angles made with the axes.

Problem 2:

Find the direction cosines of the line passing through the points A(1,2,−3)A(1, 2, -3) and B(−1,−2,1)B(-1, -2, 1).

Solution:

Step 1: Find the direction ratios (DRs) of the line segment ABAB using the formula (x2−x1,y2−y1,z2−z1)(x_2-x_1, y_2-y_1, z_2-z_1): a=−1−1=−2,b=−2−2=−4,c=1−(−3)=4a = -1 - 1 = -2, b = -2 - 2 = -4, c = 1 - (-3) = 4 Step 2: Calculate the magnitude of the segment ABAB: Magnitude=(−2)2+(−4)2+42=4+16+16=36=6\text{Magnitude} = \sqrt{(-2)^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 Step 3: Calculate the direction cosines: l=−26=−13,m=−46=−23,n=46=23l = \frac{-2}{6} = -\frac{1}{3}, m = \frac{-4}{6} = -\frac{2}{3}, n = \frac{4}{6} = \frac{2}{3} The direction cosines are (−13,−23,23)(-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}) or (13,23,−23)(\frac{1}{3}, \frac{2}{3}, -\frac{2}{3}) depending on the direction of the line.

Explanation:

For a line passing through two points, we first determine the direction ratios by subtracting the coordinates. We then normalize these ratios by dividing by the distance between the two points to obtain the direction cosines.