(A) Given direction ratios are $a = 1$,$b = -3$,and $c = 2$. The magnitude of the vector is $\sqrt{a^2 + b^2 + c^2} = \sqrt{1^2 + (-3)^2 + 2^2} = \sqr

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$\frac{1}{\sqrt{14}}, \frac{-3}{\sqrt{14}}, \frac{2}{\sqrt{14}}$

(A) Given direction ratios are $a = 1$,$b = -3$,and $c = 2$. The magnitude of the vector is $\sqrt{a^2 + b^2 + c^2} = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$. The direction cosines $(l, m, n)$ are given by $\frac{a}{\sqrt{a^2 + b^2 + c^2}}, \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \frac{c}{\sqrt{a^2 + b^2 + c^2}}$. Substituting the values,we get $\left( \frac{1}{\sqrt{14}}, \frac{-3}{\sqrt{14}}, \frac{2}{\sqrt{14}} \right)$.

Class 12 Mathematics Three Dimensional Geometry System of co-ordinates, Direction cosines and direction ratios, Projection

Easy

If the direction ratios of a line are $1, -3, 2$,then the direction cosines of the line are

A
$\frac{1}{\sqrt{14}}, \frac{-3}{\sqrt{14}}, \frac{2}{\sqrt{14}}$
B
$\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$
C
$\frac{-1}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-2}{\sqrt{14}}$
D
$\frac{-1}{\sqrt{14}}, \frac{-2}{\sqrt{14}}, \frac{-3}{\sqrt{14}}$

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System of co-ordinates, Direction cosines and direction ratios, Projection

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