Question

A vector $\mathbf { n }$ of magnitude 8 units is inclined to x-axis at $45 ^ { \circ }$ , y-axis at $60 ^ { \circ }$ and an acute angle with z-axis. If a plane passes through a point $( \sqrt { 2 } , - 1,1 )$ and is normal to $\mathbf { n }$ , then its equation in vector form is

• A. $\mathbf { r } . ( \sqrt { 2 } \mathbf { i } + \mathbf { j } + \mathbf { k } ) = 4$
• B. $\mathbf { r } . ( \sqrt { 2 } \mathbf { i } + \mathbf { j } + \mathbf { k } ) = 2$
• C. $\mathbf { r } . ( \mathbf { i } + \mathbf { j } + \mathbf { k } ) = 4$
• D. None of these

Answer 1

Answer: (B)

$\mathbf { r } . ( \sqrt { 2 } \mathbf { i } + \mathbf { j } + \mathbf { k } ) = 2$

Answer 2

Let $\gamma$be the angle made by $\mathbf { n }$ with z-axis.

Then direction cosines of $\mathbf { n }$ are $l = \cos 45 ^ { \circ } = \frac { 1 } { \sqrt { 2 } }$

$m = \cos 60 ^ { \circ } = \frac { 1 } { 2 }$ and $n = \cos \gamma$.

∴ $l ^ { 2 } + m ^ { 2 } + n ^ { 2 } = 1 \Rightarrow \left( \frac { 1 } { \sqrt { 2 } } \right) ^ { 2 } + \left( \frac { 1 } { 2 } \right) ^ { 2 } + n ^ { 2 } = 1$

⇒ $n ^ { 2 } = \frac { 1 } { 4 } \Rightarrow n = \frac { 1 } { 2 }$, [$\bullet \bullet$ $\gamma$is acute, ∴ $n = \cos \gamma > 0$]

We have $| \mathbf { n } | = 8$, ∴

$\Rightarrow \mathbf { n } = 8 \left( \frac { 1 } { \sqrt { 2 } } \mathbf { i } + \frac { 1 } { 2 } \mathbf { j } + \frac { 1 } { 2 } \mathbf { k } \right)$ $= 4 \sqrt { 2 } \mathbf { i } + 4 \mathbf { j } + 4 \mathbf { k }$

The required plane passes through the point $( \sqrt { 2 } , - 1,1 )$ having position vector $\mathbf { a } = \sqrt { 2 } \mathbf { i } - \mathbf { j } + \mathbf { k }$ .

So, its vector equation is $( \mathbf { r } - \mathbf { a } ) \cdot \mathbf { n } = 0$ or

⇒

⇒ r. $( \sqrt { 2 } \mathbf { i } + \mathbf { j } + \mathbf { k } ) = 2$.