Question ID: #854
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
- (1) 11
- (2) 3
- (3) 0
- (4) 1
Solution:
First, we find vector $\vec{c} = \vec{a} \times \vec{b}$:
$$\vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix}$$
$$\vec{c} = \hat{i}(0 – (-2)) – \hat{j}(0 – (-2)) + \hat{k}(2 – 1)$$
$$\vec{c} = 2\hat{i} – 2\hat{j} + \hat{k}$$
Calculate the magnitude of $\vec{c}$:
$$|\vec{c}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$$
Given $|\vec{c} \times \vec{d}| = 3$ and the angle between $\vec{c}$ and $\vec{d}$ is $\theta = \frac{\pi}{4}$.
$$|\vec{c} \times \vec{d}| = |\vec{c}| |\vec{d}| \sin \theta$$
$$3 = 3 \cdot |\vec{d}| \cdot \sin\left(\frac{\pi}{4}\right)$$
$$1 = |\vec{d}| \cdot \frac{1}{\sqrt{2}} \Rightarrow |\vec{d}| = \sqrt{2}$$
Given $|\vec{d} – \vec{a}| = \sqrt{11}$. Squaring both sides:
$$|\vec{d} – \vec{a}|^2 = 11$$
$$|\vec{d}|^2 + |\vec{a}|^2 – 2(\vec{a} \cdot \vec{d}) = 11$$
Calculate $|\vec{a}|^2$:
$$|\vec{a}|^2 = 2^2 + 1^2 + (-2)^2 = 4 + 1 + 4 = 9$$
Substitute the known values into the equation:
$$(\sqrt{2})^2 + 9 – 2(\vec{a} \cdot \vec{d}) = 11$$
$$2 + 9 – 2(\vec{a} \cdot \vec{d}) = 11$$
$$11 – 2(\vec{a} \cdot \vec{d}) = 11$$
$$-2(\vec{a} \cdot \vec{d}) = 0 \Rightarrow \vec{a} \cdot \vec{d} = 0$$
Ans. (3)
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