JEE Mains · Maths · STD 12 - 10. vector algebra
Let \( \vec{a}=2\hat{i}+\hat{j}-2\hat{k} \), \( \vec{b}=\hat{i}+\hat{j} \) and \( \vec{c}=\vec{a}\times \vec{b} \). Let \( \vec{d} \) be a vector such that \( {|\vec{d}-\vec{a}|}=\sqrt{11} \), \( {|\vec{c}\times\vec{d}|}=3 \) and the angle between \( \vec{c} \) and \( \vec{d} \) is \( \frac{\pi}{4} \). Then \( \vec{a}\cdot\vec{d} \) is equal to
- A 11
- B 3
- C 0
- D 1
Answer & Solution
Correct Answer
(C) 0
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ c }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|\) \(\overrightarrow{ c }=2 \hat{ i }+2 \hat{ k }+\hat{ k },| c |=3\) \({|\overrightarrow{ c } \times \overrightarrow{ d }|}=3\)…
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