First slide

Scalar triple product of vectors

Question

Given $\vec{a} = x \hat{i} + y \hat{j} + 2 \hat{k}, \vec{b} = \hat{i} - \hat{j} + \hat{k}, \vec{c} = \hat{i} + 2 \hat{j}; \vec{a} ⊥ \vec{b}, \vec{a} \cdot \vec{c} = 4$ Then

Moderate

A

$[\vec{a} \vec{b} \vec{c}]^{2} = | \vec{a} |$
B

$[\vec{a} \vec{b} \vec{c}] = | \vec{a} |$
C

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

$[\vec{a} \vec{b} \vec{c}] = | \vec{a} |^{2}$

Solution

$\vec{a} ⊥ \vec{b} \Rightarrow x - y + 2 = 0$
$\vec{a} \cdot \vec{c} = 4 \Rightarrow x + 2 y = 4$
Solving we get x=0; y=2
$\begin{array}{l} \Rightarrow \vec{a} = 2 \hat{j} + 2 \hat{k} \\ \Rightarrow [\vec{a} \vec{b} \vec{c}] = |\begin{array}{ccc} 0 & 2 & 2 \\ 1 & - 1 & 1 \\ 1 & 2 & 0 \end{array}| = 8 = | \vec{a} |^{2} \end{array}$

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