JEE Mains · Maths · STD 12 - 10. vector algebra
Let \( \vec{a}=2\hat{i}-\hat{j}-\hat{k}, \vec{b}=\hat{i}+3\hat{j}-\hat{k} \) and \( \vec{c}=2\hat{i}+\hat{j}+3\hat{k} \). Let \( \vec{v} \) be the vector in the plane of the vectors \( \vec{a} \) and \( \vec{b} \), such that the length of its projection on the vector \( \vec{c} \) is equal to \( \frac{1}{\sqrt{14}} \). Then \( |\vec{v}| \) is equal is
- A \( \frac{\sqrt{21}}{2} \)
- B 13
- C \( \frac{\sqrt{35}}{2} \)
- D 7
Answer & Solution
Correct Answer
(C) \( \frac{\sqrt{35}}{2} \)
Step-by-step Solution
Detailed explanation
\( \vec{v}=x\vec{a}+y\vec{b}=x(2\hat{i}-\hat{j}-\hat{k})+y(\hat{i}+3\hat{j}-\hat{k}) \) \( \vec{v}=(2x+y)\hat{i}+(3y-x)\hat{j}+(-x-y)\hat{k} \) \(\left|\frac{\vec{ v } \cdot \vec{ c }}{|\vec{ c }|}\right|=\frac{1}{\sqrt{14}}\) \(\vec{v} \cdot \vec{c}=2(2 x+y)+3 y-x-3 x-3 y\)…
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