JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec{a}=2 \hat i-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\). Let a vector \(\overrightarrow{\mathrm{v}}\) be in the plane containing \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{b}}\). If \(\overrightarrow{\mathrm{v}}\) is perpendicular to the vector \(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\) and its projection on \(\vec{a}\) is \(19\, units,\) then \(|2 \vec{v}|^{2}\) is equal to .... .
- A \(1400\)
- B \(149\)
- C \(494\)
- D \(1494\)
Answer & Solution
Correct Answer
(D) \(1494\)
Step-by-step Solution
Detailed explanation
\(\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}\) \(\vec{b}=\hat{i}+2 \hat{j}-\hat{k}\) \(\vec{c}=3 \hat{i}+2 \hat{j}-\hat{k}\) \(\vec{v}=x \vec{a}+y \vec{b} \quad \vec{v}(3 \hat{i}+2 \hat{j}-k)=0\) \(\vec{v} \cdot \hat{a}=19\) \(\vec{v}=\lambda \vec{c} \times(\vec{a} \times \vec{b})\)…
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