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Vectors - Direction Cosines and Direction Ratios

Grade 12ICSE

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

🔑Concepts

Direction Cosines (DCs) are the cosines of the angles α\alpha, β\beta, and γ\gamma that a vector makes with the positive directions of the xx, yy, and zz axes respectively. Visually, if you imagine a vector r\vec{r} originating from the origin OO to a point PP, α\alpha is the angle between the vector and the xx-axis, β\beta with the yy-axis, and γ\gamma with the zz-axis.

Direction Cosines are typically denoted by the letters ll, mm, and nn, where l=cosαl = \cos \alpha, m=cosβm = \cos \beta, and n=cosγn = \cos \gamma. A key geometric property is that these values represent the coordinates of a point on a unit sphere centered at the origin, meaning the magnitude of the vector (l,m,n)(l, m, n) is always 11.

Direction Ratios (DRs) are any three numbers a,b,ca, b, c that are proportional to the direction cosines l,m,nl, m, n. While a vector has a unique set of direction cosines, it can have infinitely many sets of direction ratios. Visually, any vector parallel to the original vector will share the same direction ratios, differing only by a scalar multiple kk.

The sum of the squares of the direction cosines of a line is always equal to 11, expressed as l2+m2+n2=1l^2 + m^2 + n^2 = 1. This implies that cos2α+cos2β+cos2γ=1\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1. This identity is crucial for finding a missing angle or cosine value when the other two are known.

If a vector is given as r=ai^+bj^+ck^\vec{r} = a\hat{i} + b\hat{j} + c\hat{k}, the components a,b,ca, b, c are the direction ratios of the vector. To visualize this, these components represent the steps taken along the three primary axes to reach the terminal point of the vector from its starting point.

The direction cosines of a line segment joining 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) are found by first determining the direction ratios (x2x1),(y2y1),(z2z1)(x_2 - x_1), (y_2 - y_1), (z_2 - z_1), and then dividing each by the distance PQPQ. Visually, this normalizes the displacement vector between the two points to a unit length.

Two lines with direction ratios (a1,b1,c1)(a_1, b_1, c_1) and (a2,b2,c2)(a_2, b_2, c_2) are parallel if a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}. They are perpendicular if the sum of the products of their corresponding ratios is zero, i.e., a1a2+b1b2+c1c2=0a_1 a_2 + b_1 b_2 + c_1 c_2 = 0. This dot-product relationship is fundamental in 3D geometry.

📐Formulae

l=cosα,m=cosβ,n=cosγl = \cos \alpha, m = \cos \beta, n = \cos \gamma

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

cos2α+cos2β+cos2γ=1\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1

sin2α+sin2β+sin2γ=2\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 2

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

Direction Ratios of PQ=(x2x1,y2y1,z2z1)\text{Direction Ratios of } \vec{PQ} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)

\text{Distance } PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

cosθ=l1l2+m1m2+n1n2\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|

💡Examples

Problem 1:

A line makes angles 9090^\circ, 6060^\circ, and 3030^\circ with the positive xx, yy, and zz axes respectively. Find its direction cosines.

Solution:

  1. Identify the angles: α=90,β=60,γ=30\alpha = 90^\circ, \beta = 60^\circ, \gamma = 30^\circ.
  2. Calculate l=cos90=0l = \cos 90^\circ = 0.
  3. Calculate m=cos60=12m = \cos 60^\circ = \frac{1}{2}.
  4. Calculate n=cos30=32n = \cos 30^\circ = \frac{\sqrt{3}}{2}.
  5. Verify the identity: l2+m2+n2=02+(12)2+(32)2=0+14+34=1l^2 + m^2 + n^2 = 0^2 + (\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2 = 0 + \frac{1}{4} + \frac{3}{4} = 1.
  6. The direction cosines are (0,12,32)(0, \frac{1}{2}, \frac{\sqrt{3}}{2}).

Explanation:

Direction cosines are simply the cosine values of the angles a vector makes with the coordinate axes. The solution confirms these values by checking if the sum of their squares equals 1.

Problem 2:

Find the direction cosines of the line passing through the points A(2,4,5)A(-2, 4, -5) and B(1,2,3)B(1, 2, 3).

Solution:

  1. Find the Direction Ratios (DRs) by subtracting coordinates: a=(1(2))=3,b=(24)=2,c=(3(5))=8a = (1 - (-2)) = 3, b = (2 - 4) = -2, c = (3 - (-5)) = 8.
  2. Calculate the magnitude of the vector ABAB: AB=32+(2)2+82=9+4+64=77|\vec{AB}| = \sqrt{3^2 + (-2)^2 + 8^2} = \sqrt{9 + 4 + 64} = \sqrt{77}.
  3. Calculate DCs using the formula l=aAB,m=bAB,n=cABl = \frac{a}{|\vec{AB}|}, m = \frac{b}{|\vec{AB}|}, n = \frac{c}{|\vec{AB}|}.
  4. l=377,m=277,n=877l = \frac{3}{\sqrt{77}}, m = \frac{-2}{\sqrt{77}}, n = \frac{8}{\sqrt{77}}.
  5. The direction cosines are (377,277,877)(\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}).

Explanation:

To find the direction cosines of a line segment, we first find the direction ratios by taking the difference of the coordinates. Then, we normalize these ratios by dividing them by the total distance between the two points.