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MHT CET · Maths · Vector Algebra

For non-zero vectors \(\bar{a}, \bar{b}, \bar{c},|(\bar{a} \times \bar{b}) \cdot \bar{c}|=|\bar{a}||\bar{b}||\bar{c}|\) holds and only if

  1. A \(\bar{b} \cdot \bar{c}=0, \bar{c} \cdot \bar{a}=0\)
  2. B \(\bar{c} \cdot \bar{a}=0, \bar{a} \cdot \bar{b}=0\)
  3. C \(\bar{a} \cdot \bar{b}=\bar{b} \cdot \bar{c}=\bar{c} \cdot \bar{a}=0\)
  4. D \(\bar{a} \times \bar{b}=0, \bar{b} \times \bar{c}=0\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\bar{a} \cdot \bar{b}=\bar{b} \cdot \bar{c}=\bar{c} \cdot \bar{a}=0\)

Step-by-step Solution

Detailed explanation

\(|(\vec{a} \times \vec{b}) \cdot \vec{c}| \)
\( =|\vec{a}||\vec{b}||\vec{c}| \)
\( \Rightarrow|\vec{a} \times \vec{b}||\vec{c}| \cos \theta=|\vec{a}||\vec{b}||\vec{c}| \)
\( \Rightarrow|\vec{a}||\vec{b}| \sin \phi|\vec{c}| \cos \theta=|\vec{a}||\vec{b}||\vec{c}| \)
\( \Rightarrow \sin \phi=1 \text { and } \cos \theta=1 \)
\( \phi=90^{\circ} \text { and } \theta=0^{\circ} \)
\( \text {where } \phi \text { is the angle between } \vec{a} \text { and } \vec{b} \text { and } \theta \text { is the angle}\) \(\text{between } \vec{c} \text { and } \)
\( \text {normal to the plane containing } \vec{a} \text { and } \vec{b} \)
\( \Rightarrow \vec{a} \perp \vec{b}, \vec{b} \perp \vec{c} \text {, and } \vec{c} \perp \vec{a} \)
\( \Rightarrow \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0\)