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Question: \(|(\mathbf{a} \times \mathbf{b}).\mathbf{c}| = |\mathbf{a}||\mathbf{b}||\mathbf{c}|,\) if...

∣(a×b).c∣=∣a∣∣b∣∣c∣,|(\mathbf{a} \times \mathbf{b}).\mathbf{c}| = |\mathbf{a}||\mathbf{b}||\mathbf{c}|, if

A

a.b=b.c=0\mathbf{a}.\mathbf{b} = \mathbf{b}.\mathbf{c} = 0

B

b.c=c.a=0\mathbf{b}.\mathbf{c} = \mathbf{c}.\mathbf{a} = 0

C

c.a=a.b=0\mathbf{c}.\mathbf{a} = \mathbf{a}.\mathbf{b} = 0

D

a.b=b.c=c.a=0\mathbf{a}.\mathbf{b} = \mathbf{b}.\mathbf{c} = \mathbf{c}.\mathbf{a} = 0

Answer

a.b=b.c=c.a=0\mathbf{a}.\mathbf{b} = \mathbf{b}.\mathbf{c} = \mathbf{c}.\mathbf{a} = 0

Explanation

Solution

We have ∣(a×b).c∣=∣a∣∣b∣∣c∣|(\mathbf{a} \times \mathbf{b}).\mathbf{c}| = |\mathbf{a}||\mathbf{b}||\mathbf{c}|

⇒∣∣a∣∣b∣sin⁡θn.c∣=∣a∣∣b∣∣c∣\Rightarrow \left| |\mathbf{a}||\mathbf{b}|\sin\theta\mathbf{n}.\mathbf{c} \right| = |\mathbf{a}||\mathbf{b}||\mathbf{c}|

⇒∣∣a∣∣b∣∣c∣sin⁡θcos⁡α∣=∣a∣∣b∣∣c∣\Rightarrow \left| |\mathbf{a}||\mathbf{b}||\mathbf{c}|\sin\theta\cos\alpha \right| = |\mathbf{a}||\mathbf{b}||\mathbf{c}|

⇒ ∣sin⁡θ∣∣cos⁡α∣=1⇒θ=π2\Rightarrow \ |\sin\theta||\cos\alpha| = 1 \Rightarrow \theta = \frac{\pi}{2} and α=0\alpha = 0

⇒a⊥b\Rightarrow \mathbf{a}\bot\mathbf{b} and c∣∣n\mathbf{c}||\mathbf{n}

⇒a⊥b\Rightarrow \mathbf{a}\bot\mathbf{b} and c\mathbf{c}is perpendicular to both a\mathbf{a}and b\mathbf{b}

∴ a,b,c\mathbf{a},\mathbf{b},\mathbf{c} are mutually perpendicular

Hence, a.b=b.c=c.a=0.\mathbf{a}.\mathbf{b} = \mathbf{b}.\mathbf{c} = \mathbf{c}.\mathbf{a} = 0.