JEE MAIN - Mathematics (2023 - 13th April Morning Shift - No. 3)
Let $$\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$$ and $$\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$$. If a vector $$\vec{d}$$ satisfies $$\vec{d} \times \vec{b}=\vec{c} \times \vec{b}$$ and $$\vec{d} \cdot \vec{a}=24$$, then $$|\vec{d}|^{2}$$ is equal to :
313
413
423
323
Explanation
Given that $$\vec{d} \times \vec{b} = \vec{c} \times \vec{b}$$, we can rewrite this as:
$$(\vec{d} - \vec{c}) \times \vec{b} = \vec{0}$$
This implies that the vector $$\vec{d} - \vec{c}$$ is a scalar multiple of $$\vec{b}$$:
$$\vec{d} = \vec{c} + \lambda \vec{b}$$
Also, we are given that $$\vec{d} \cdot \vec{a} = 24$$:
$$(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 24$$
Now, we can find the value of $$\lambda$$ :
$$\lambda = \frac{24 - \vec{a} \cdot \vec{c}}{\vec{b} \cdot \vec{a}} = \frac{24 - 6}{9} = 2$$
Therefore, we have :
$$\vec{d} = \vec{c} + 2(\vec{b}) = 8\hat{i} - 5\hat{j} + 18\hat{k}$$
Now, we can find the squared magnitude of $$\vec{d}$$ :
$$|\vec{d}|^2 = 64 + 25 + 324 = 413$$
$$(\vec{d} - \vec{c}) \times \vec{b} = \vec{0}$$
This implies that the vector $$\vec{d} - \vec{c}$$ is a scalar multiple of $$\vec{b}$$:
$$\vec{d} = \vec{c} + \lambda \vec{b}$$
Also, we are given that $$\vec{d} \cdot \vec{a} = 24$$:
$$(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 24$$
Now, we can find the value of $$\lambda$$ :
$$\lambda = \frac{24 - \vec{a} \cdot \vec{c}}{\vec{b} \cdot \vec{a}} = \frac{24 - 6}{9} = 2$$
Therefore, we have :
$$\vec{d} = \vec{c} + 2(\vec{b}) = 8\hat{i} - 5\hat{j} + 18\hat{k}$$
Now, we can find the squared magnitude of $$\vec{d}$$ :
$$|\vec{d}|^2 = 64 + 25 + 324 = 413$$
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