Let $\mathbf{b}=-\hat{\mathbf{i}}+4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$ and $\mathbf{c}=2 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}$. If $\mathbf{a}$ be $a$ unit vector and the scalar triple product $[\mathbf{a} \mathbf{b} \mathbf{c}]$ has the greatest value, then a is equal to

$\frac{1}{\sqrt{3}}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})$

$\frac{1}{\sqrt{5}}(\sqrt{2} \hat{i}-\hat{j}-\sqrt{2} \hat{k})$

$\frac{1}{\sqrt{59}}(3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-\hat{\mathbf{k}})$

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

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Application +4 | -1

Let

\mathbf{b}=-\hat{\mathbf{i}}+4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}

and

\mathbf{c}=2 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}

. If

\mathbf{a}

a

unit vector and the scalar triple product

[\mathbf{a} \mathbf{b} \mathbf{c}]

has the greatest value, then a is equal to

View answer and explanation

Correct option: C

\frac{1}{3}(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})

Option C (

\frac{1}{3}(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})

) is correct because the scalar triple product is maximized when a is in the direction of b x c, and normalizing b x c gives the unit vector a.