Question

A non-zero vector a is parallel to the line of intersection of the plane determined by the vectors i, i + j and the plane determined by the vectors i – j, i + k. The angle between a and the vector i – 2j + 2k is

• A. $\frac { \pi } { 4 }$ or$\frac { 3 \pi } { 4 }$
• B. $\frac { 2 \pi } { 4 }$ or $\frac { 3 \pi } { 4 }$
• C. $\frac { \pi } { 2 }$or $\frac { 3 \pi } { 2 }$
• D. None of these

Answer 1

Answer: (A)

$\frac { \pi } { 4 }$ or$\frac { 3 \pi } { 4 }$

Answer 2

Equation of plane containing i and i + j is

$[ r - i , i , i + j ] = 0$

⇒

⇒ ⇒ z = 0 ……(i)

Equation of plane containing i – j and i + k is

⇒ $\left[ \begin{array} { l l l } { [ \mathbf { r } - ( \mathbf { i } - \mathbf { j } ) } & \mathbf { i } - \mathbf { j } & \mathbf { i } + \mathbf { k } ] = 0 \end{array} \right.$

⇒ $( \mathbf { r } - \mathbf { i } + \mathbf { j } ) [ ( \mathbf { i } - \mathbf { j } ) \times ( \mathbf { i } + \mathbf { k } ) ] = 0$

⇒ $x + y - z = 0$ …… (ii)

Let .

Since a is parallel to (i) and (ii)

$a _ { 3 } = 0$, $a _ { 1 } + a _ { 2 } - a _ { 3 } = 0$

⇒ $a _ { 1 } = - a _ { 2 }$, $a _ { 3 } = 0$

Thus a vector in the direction of a is .

If θ is the angle between a and .

Then $\cos \theta = \pm \frac { 1 ( 1 ) + ( - 1 ) ( - 2 ) } { \sqrt { 1 + 1 } \sqrt { 1 + 4 + 4 } } = \pm \frac { 3 } { \sqrt { 2 } \cdot 3 }$

⇒ $\cos \theta = \pm \frac { 1 } { \sqrt { 2 } }$

⇒ $\theta = \pi / 4$ or $3 \pi / 4$