Question

The value of b such that scalar product of the vectors $( \mathbf { i } + \mathbf { j } + \mathbf { k } )$ with the unit vector parallel to the sum of the vectors $( 2 \mathbf { i } + 4 \mathbf { j } - 5 \mathbf { k } )$ and $( b \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } )$ is 1, is

• A. – 2
• B. – 1
• C. 0
• D. 1

Answer 1

Answer: (D)

1

Answer 2

Parallel vector $= ( 2 + b ) \mathbf { i } + 6 \mathbf { j } - 2 \mathbf { k }$

Unit vector $= \frac { ( 2 + b ) \mathbf { i } + 6 \mathbf { j } - 2 \mathbf { k } } { \sqrt { b ^ { 2 } + 4 b + 44 } }$

According to the condition, $1 = \frac { ( 2 + b ) + 6 - 2 } { \sqrt { b ^ { 2 } + 4 b + 44 } }$

$\Rightarrow b ^ { 2 } + 4 b + 44 = b ^ { 2 } + 12 b + 36$ $\Rightarrow 8 b = 8 \Rightarrow b = 1$