Let a = - i + j + 2 k , b = i - j - 3 k , c = a × b and d = c × a . Then ( a - b ) · d is equal to: [2026]

Question

Let $\vec{a} = - \hat{i} + \hat{j} + 2 \hat{k},$ $\vec{b} = \hat{i} - \hat{j} - 3 \hat{k},$ $\vec{c} = \vec{a} \times \vec{b}$ and $\vec{d} = \vec{c} \times \vec{a} .$ Then $(\vec{a} - \vec{b}) \cdot \vec{d}$ is equal to: [2026]

Smart Achievers · Accepted Answer

(2)

$\vec{d} = (\vec{a} \times \vec{b}) \times \vec{a}$

$\vec{d} = (a^{2}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{a}$

$\vec{d} = 6 \vec{b} + 8 \vec{a}$

$(\vec{a} - \vec{b}) \cdot \vec{d} = (\vec{a} - \vec{b}) \cdot (6 \vec{b} + 8 \vec{a})$

$= 8 a^{2} - 6 b^{2} - 2 \vec{a} \cdot \vec{b}$

$= 48 - 66 + 16 = - 2$

Solution

Q.
Let $\vec{a} = - \hat{i} + \hat{j} + 2 \hat{k},$ $\vec{b} = \hat{i} - \hat{j} - 3 \hat{k},$ $\vec{c} = \vec{a} \times \vec{b}$ and $\vec{d} = \vec{c} \times \vec{a} .$ Then $(\vec{a} - \vec{b}) \cdot \vec{d}$ is equal to: [2026]

1 $- 4$

2 $- 2$

3 $4$

4 $2$

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Solution

Q. Let a→=-i^+j^+2k^, b→=i^-j^-3k^, c→=a→×b→ and d→=c→×a→. Then (a→-b→)·d→ is equal to: [2026]

1 -4

2 -2

3 4

4 2