MHT CET · Maths · Vector Algebra
If \(\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}\) are mutually perpendicular vectors having magnitudes \(1,2,3\) respectively, then the value of \(\left[\begin{array}{lll}\bar{a}+\bar{b}+\bar{c} & \bar{b}-\bar{a} & \bar{c}\end{array}\right]\) is
- A 0
- B 6
- C 12
- D 18
Answer & Solution
Correct Answer
(C) 12
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & |\overline{\mathrm{a}}|=1,|\overline{\mathrm{~b}}|=2,|\overline{\mathrm{c}}|=3 \\ & {[\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}} \overline{\mathrm{b}}-\overline{\mathrm{a}} \overline{\mathrm{c}}]} \\ & =(\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}) \cdot[(\overline{\mathrm{b}}-\overline{\mathrm{a}}) \times \overline{\mathrm{c}}] \\ & =(\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}) \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}}-\overline{\mathrm{a}} \times \overline{\mathrm{c}}) \\ & =[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]-[\overline{\mathrm{b}} \overline{\mathrm{a}} \overline{\mathrm{c}}] \\ & =2[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}] \\ & =2 \overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \\ & =2|\overline{\mathrm{a}}||\overline{\mathrm{b}} \times \overline{\mathrm{c}}| \cos 0^{\circ} \\ & =2|\overline{\mathrm{a}}||\overline{\mathrm{b}} \times \overline{\mathrm{c}}| \\ & =2|\overline{\mathrm{a}}||\overline{\mathrm{b}}||\overline{\mathrm{c}}| \sin 90^{\circ} \\ & =2 \times 1 \times 2 \times 3 \\ & =12\end{aligned}\)
\(\begin{aligned} & \cdots\left[\begin{array}{l}\because \overline{\mathrm{a}} \perp \overline{\mathrm{b}} \text { and } \overline{\mathrm{c}} \\ \therefore \overline{\mathrm{a}} \| \overline{\mathrm{b}} \times \overline{\mathrm{c}}\end{array}\right] \\ & \cdots[\because \overline{\mathrm{b}} \perp \overline{\mathrm{c}}]\end{aligned}\)
\(\begin{aligned} & \cdots\left[\begin{array}{l}\because \overline{\mathrm{a}} \perp \overline{\mathrm{b}} \text { and } \overline{\mathrm{c}} \\ \therefore \overline{\mathrm{a}} \| \overline{\mathrm{b}} \times \overline{\mathrm{c}}\end{array}\right] \\ & \cdots[\because \overline{\mathrm{b}} \perp \overline{\mathrm{c}}]\end{aligned}\)
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