Question

Direction cosine of normal to plane containing

x = y = z and x –1 = y –1 =$\frac{z - 1}{d}$; where

dĪ R ~ {1} are-

• A. $\left\{ \frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}} \right\}$
• B. $\left\{ - \frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}} \right\}$
• C. $\left\{ \frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}},0 \right\}$
• D. None of these

Answer 1

Answer: (C)

$\left\{ \frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}},0 \right\}$

Answer 2

Let l, m, n be direction ratio of normal

\ l + m + n = 0 …(i)

l + m + nd = 0 …(ii)

Solve n (d –1) = 0

n = 0

\ l = ±$\frac{1}{\sqrt{2}}$

m = $\mp \frac{1}{\sqrt{2}}$

\ $\left( \frac{1}{\sqrt{2}},\frac{- 1}{\sqrt{2}},0 \right)$ or $\left( \frac{- 1}{\sqrt{2}},\frac{1}{\sqrt{2}},0 \right)$