JEE Mains · Maths · STD 12 - 10. vector algebra
Let a vector \(\vec{a}\) be coplanar with vectors \(\vec{b}=2 \hat{i}+\hat{j}+\hat{k}\) and \(\vec{c}=\hat{i}-\hat{j}+\hat{k} .\) If \(\vec{a}\) is perpendicular to \(\vec{d}=3 \vec{i}+2 \hat{j}+6 \hat{k}\), and \(|\vec{a}|=\sqrt{10} .\) Then a possible value of \([\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]+[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \vec{d}]+[\overrightarrow{\mathrm{a}} \vec{c} \vec{d}]\) is equal to:
- A \(-40\)
- B \(-42\)
- C \(-29\)
- D \(-38\)
Answer & Solution
Correct Answer
(B) \(-42\)
Step-by-step Solution
Detailed explanation
\(\vec{a}=\lambda \vec{b}+\mu \vec{c}=\hat{i}(2 \lambda+\mu)+\hat{j}(\lambda-\mu)+\hat{k}(\lambda+\mu)\) \(\vec{a} \cdot \vec{d}=0=3(2 \lambda+\mu)+2(\lambda-\mu)+6(\lambda+\mu)\) \(\Rightarrow 14 \lambda+7 \mu=0 \Rightarrow \mu=-2 \lambda\)…
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