MHT CET · Maths · Vector Algebra
If \(\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}\) are non-coplanar vectors and \(\overline{\mathrm{p}}=\frac{\overline{\mathrm{b}} \times \overline{\mathrm{c}}}{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]}, \overline{\mathrm{q}}=\frac{\overline{\mathrm{c}} \times \overline{\mathrm{a}}}{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]}, \overline{\mathrm{r}}=\frac{\overline{\mathrm{a}} \times \overline{\mathrm{b}}}{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]}, \quad\) then \(2 \overline{\mathrm{a}} \cdot \overline{\mathrm{p}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{q}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{r}}=\)
- A 0
- B 3
- C 4
- D 1
Answer & Solution
Correct Answer
(C) 4
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& 2 \overline{\mathrm{a}} \cdot \overline{\mathrm{p}}=\frac{2(\overline{\mathrm{~b}} \times \overline{\mathrm{c}}) \cdot \overline{\mathrm{a}}}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}=\frac{2[\overline{\mathrm{~b}} \overline{\mathrm{c}} \mathrm{a}]}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}=2 . \\
& \overline{\mathrm{b}} \cdot \overline{\mathrm{q}}=\frac{(\overline{\mathrm{c}} \times \overline{\mathrm{a}}) \cdot \overline{\mathrm{b}}}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}=\frac{[\overline{\mathrm{c}} \overline{\mathrm{a}} \overline{\mathrm{~b}}]}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}=1 \\
& \overline{\mathrm{c}} \cdot \overline{\mathrm{r}}=\frac{(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \cdot \overline{\mathrm{c}}}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}=\frac{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}=1 \\
& \therefore \quad 2 \overline{\mathrm{a}} \cdot \overline{\mathrm{p}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{q}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{r}}=2+1+1=4
\end{aligned}\)
& 2 \overline{\mathrm{a}} \cdot \overline{\mathrm{p}}=\frac{2(\overline{\mathrm{~b}} \times \overline{\mathrm{c}}) \cdot \overline{\mathrm{a}}}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}=\frac{2[\overline{\mathrm{~b}} \overline{\mathrm{c}} \mathrm{a}]}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}=2 . \\
& \overline{\mathrm{b}} \cdot \overline{\mathrm{q}}=\frac{(\overline{\mathrm{c}} \times \overline{\mathrm{a}}) \cdot \overline{\mathrm{b}}}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}=\frac{[\overline{\mathrm{c}} \overline{\mathrm{a}} \overline{\mathrm{~b}}]}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}=1 \\
& \overline{\mathrm{c}} \cdot \overline{\mathrm{r}}=\frac{(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \cdot \overline{\mathrm{c}}}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}=\frac{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}=1 \\
& \therefore \quad 2 \overline{\mathrm{a}} \cdot \overline{\mathrm{p}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{q}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{r}}=2+1+1=4
\end{aligned}\)
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