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Vector Algebra Part 2

Section: Mathematics

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Question : 80 of 100

Marks: +1, -0

Let a vector

\vec{a}

be coplanar with vectors

\vec{b}=2\hat{i}+\hat{j}+\hat{k}

\vec{c}=\hat{i}-\hat{j}+\hat{k}

\vec{a}

is perpendicular to

\vec{d}=3\hat{i}+2\hat{j}+6\hat{k}

|\vec{a}|=\sqrt{10}

. Then a possible value of

[\vec{a}\vec{b}\vec{c}]+[\vec{a}\vec{b}\vec{d}]+[\vec{a}\vec{c}\vec{d}]

[22 Jul 2021 Shift 2]

$-42$
$-40$
$-29$
$-38$

Solution:

\vec{a}=\lambda\vec{b}+\mu\vec{c}=\hat{i}(2\lambda+\mu)+\hat{j}(\lambda-\mu)+\hat{k}(\lambda+\mu)

\vec{a}\cdot\vec{d}=0=3(2\lambda+\mu)+2(\lambda-\mu)+6(\lambda+\mu)

\Rightarrow 14\lambda+7\mu=0 \Rightarrow \mu=-2\lambda

\Rightarrow \vec{a}=(0)\hat{i}-3\lambda\hat{j}+(-\lambda)\hat{k}

\Rightarrow |\vec{a}|=\sqrt{10}|\lambda|=\sqrt{10} \Rightarrow |\lambda|=1

\Rightarrow \lambda=1

or

-1

[\vec{a}\vec{b}\vec{c}]=0

[\vec{a}\vec{b}\vec{c}]+[\vec{a}\vec{b}\vec{d}]+[\vec{a}\vec{c}\vec{d}]

=[\vec{a}\vec{b}+\vec{c}\vec{d}]

=\begin{vmatrix} 0 & -3\lambda & \lambda \\ 3 & 0 & 2 \\ 3 & 2 & 6 \end{vmatrix}

=3\lambda(12)+\lambda(6)=42\lambda=-42

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