Question

A unit vector which is coplanar to vector $\mathbf { i } + \mathbf { j } + 2 \mathbf { k }$ and $\mathbf { i } + 2 \mathbf { j } + \mathbf { k }$ and perpendicular to $\mathbf { i } + \mathbf { j } + \mathbf { k }$ is

• A. $\frac { \mathbf { i } - \mathbf { j } } { \sqrt { 2 } }$
• B. $\pm \left( \frac { \mathbf { j } - \mathbf { k } } { \sqrt { 2 } } \right)$
• C. $\frac { \mathbf { k } - \mathbf { i } } { \sqrt { 2 } }$
• D. $\frac { \mathbf { i } + \mathbf { j } + \mathbf { k } } { \sqrt { 3 } }$

Answer 1

Answer: (B)

$\pm \left( \frac { \mathbf { j } - \mathbf { k } } { \sqrt { 2 } } \right)$

Answer 2

Let the vector be given as For this vector to be coplanar with $\mathbf { i } + \mathbf { j } + 2 \mathbf { k }$ and $\mathbf { i } + 2 \mathbf { j } + \mathbf { k }$ we will have $a \mathbf { i } + b \mathbf { j } + c \mathbf { k } = p ( \mathbf { i } + \mathbf { j } + 2 \mathbf { k } ) + r ( \mathbf { i } + 2 \mathbf { j } + \mathbf { k } )$

This gives, $a = p + r$ .....(i)

$b = p + 2 r$ .....(ii)

$c = 2 p + r$ .....(iii)

For the vector $a \mathbf { i } + b \mathbf { j } + c \mathbf { k }$ to be perpendicular to $\mathbf { i } + \mathbf { j } + \mathbf { k }$ we will have $( a \mathbf { i } + b \mathbf { j } + c \mathbf { k } ) \cdot ( \mathbf { i } + \mathbf { j } + \mathbf { k } ) = 0$

$\Rightarrow a + b + c = 0$ ......(iv)

Adding equation (i) to (iii), we get $4 p + 4 r = a + b + c$

$\Rightarrow 4 ( p + r ) = 0 \Rightarrow p = - r$

Now with the help of (i), (ii) and (iii), we get

$a = 0$ $b = r$ $c = p = - r$

Hence the required vector is $r ( \mathbf { j } - \mathbf { k } )$

To be its unit vector $r ^ { 2 } + r ^ { 2 } = 1 \Rightarrow r = \pm \frac { 1 } { \sqrt { 2 } }$

Hence the required unit vector is, $\pm \frac { 1 } { \sqrt { 2 } } ( \mathbf { j } - \mathbf { k } )$.

Trick : Check for option is a unit vector and perpendicular to $\mathbf { i } + \mathbf { j } + \mathbf { k }$.

But $\left| \begin{array} { c c c } \frac { 1 } { \sqrt { 2 } } & \frac { - 1 } { \sqrt { 2 } } & 0 \\ 1 & 1 & 2 \\ 1 & 2 & 1 \end{array} \right| = - \frac { 4 } { \sqrt { 2 } } \neq 0$.

So it is not coplanar with the given vector.

Check for option is a unit vector and also perpendicular to $\mathbf { i } + \mathbf { j } + \mathbf { k }$ $\left| \begin{array} { c c c } 0 & \frac { 1 } { \sqrt { 2 } } & \frac { - 1 } { \sqrt { 2 } } \\ 1 & 1 & 2 \\ 1 & 2 & 1 \end{array} \right| = 0$ .

So, it is also coplanar with the given vectors.