# Let → a = ^ i + 4 ^ j + 2 ^ k , → b = 3 ^ i − 2 ^ j + 7 ^ k and → c = 2 ^ i − ^ j + 4 ^ k . Find a vector → d which is perpendicular to both → a and → d and → c ⋅ → d = 15 . - Mathematics

Sum

Let $\vec{a} = \hat{ i } + 4 \hat{ j } + 2 \hat{ k } , \vec{b} = 3 \hat{ i }- 2 \hat{ j } + 7 \hat{ k } \text{ and } \vec{c} = 2 \hat{ i } - \hat{ j } + 4 \hat{ k } .$  Find a vector $\vec{d}$ which is perpendicular to both $\vec{a} \text{ and } \vec{d}$ $\text{ and } \vec{c} \cdot \vec{d} = 15 .$

#### Solution

$\text{ Given } :$
$\vec{a} = \hat{ i } + 4 \hat{ j } + 2 \hat{ k }$
$\vec{b} = 3 \hat{ i } - 2 \hat{ j } + 7 \hat{ k }$
$\vec{c} = 2 \hat{ i } - \hat{ j } + 4 \hat{ k }$
$\text{ Since d is perpendicular to both a and b, it is parallel to } \vec{a} \times \vec{b} .$
$\text{ Suppose } d = \lambda\left( \vec{a} \times \vec{b} \right) \text{ for some scalar } \lambda .$
$d = \lambda \begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & 4 & 2 \\ 3 & - 2 & 7\end{vmatrix}$
$= \lambda \left[ \left( 28 + 4 \right) \hat{ i } - \left( 7 - 6 \right) \hat{ j } + \left( - 2 - 12 \right) \hat{ k } \right]$
$= \lambda \left[ 32 \hat{ i } - \hat{ j } - 14 \hat{ k } \right]$
$\vec{c .} \vec{d} = 15 (\text{ Given } )$
$\Rightarrow \left( 2 \hat{ i } - \hat{ j } + 4 \hat{ k } \right) . \lambda \left( 32 \hat{ i }- \hat{ j } - 14 \hat{ k } \right) = 15$
$\Rightarrow \lambda\left( 64 + 1 - 56 \right) = 15$
$\Rightarrow \lambda = \frac{5}{3}$
$\therefore \vec{d} = \frac{5}{3}\left( 32 \hat{ i } - \hat{ j } - 14 \hat{ k } \right)$
$\Rightarrow \vec{d} = \frac{1}{3}\left( 160 \hat{ i } - 5 \hat{ j } - 70 \hat{ k } \right)$

#### Notes

The question should contain  $"\text{ which is perpendicular to both } \vec{a} \text{ and } \vec{b} "$
$\text{ instead of }$
$"\text { which is perpendicular to both } \vec{a} \text{ and } \vec{d} "$

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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 25 Vector or Cross Product
Exercise 25.1 | Q 27 | Page 30