First slide

Direction ratios and direction cosines

Question

The direction cosines of the line of intersection of two planes $x - y + z - 5 = 0, x - 3 y - 6 = 0$ are

Easy

A

$⟨ 3, 1, - 2 ⟩$
B

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

$⟨ 1, 0, 1 ⟩$
D

$⟨ 0, 1, 0 ⟩$

Solution

The direction ratios of line of intersection of two planes are proportional to the vector which is perpendicular to both normal vectors of the planes
The given planes are $x - y + z - 5 = 0, x - 3 y - 6 = 0$
Its normal vectors are $i - j + k, i - 3 j$
vector perpendicular to both the vectors is $|\begin{matrix} i & j & k \\ 1 & - 1 & 1 \\ 1 & - 3 & 0 \end{matrix}| = i (3) - j (- 1) + k (- 2) = 3 i - j - 2 k$
The line of intersection of two planes having direction ratios $⟨ 3, - 1, - 2 ⟩$ and its direction cosines are $⟨ \frac{3}{\sqrt{14}}, - \frac{1}{\sqrt{14}}, - \frac{2}{\sqrt{14}} ⟩$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App