Question

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. 1
• B. 3
• C. 11
• D. 0
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Hint:

Use vector identities to reduce problems involving magnitudes and angles to simple dot products.

Answer 1

The Correct Option is D

Step 1: Finding vector $\vec c$.
\[ \vec c = \vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} \] \[ \vec c = 2\hat i - 2\hat j + \hat k \] Step 2: Using the cross product magnitude condition.
\[ |\vec c \times \vec d| = |\vec c||\vec d|\sin\theta \] \[ 3 = |\vec c||\vec d|\sin\frac{\pi}{4} \] \[ |\vec c| = \sqrt{4+4+1}=3 \Rightarrow |\vec d|=\sqrt{2} \] Step 3: Using $|\vec d-\vec a|=\sqrt{11$.}
\[ |\vec d-\vec a|^2 = |\vec d|^2 + |\vec a|^2 -2\vec a\cdot\vec d \] \[ 11 = 2 + 9 - 2\vec a\cdot\vec d \] \[ \vec a\cdot\vec d = 0 \]