Vector Algebra Question 3
Question 3 - 24 January - Shift 2
Let $\vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \quad \vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}, \vec{a} \cdot \vec{c}=7$, $2 \vec{b} \cdot \vec{c}+43=0, \vec{a} \times \vec{c}=\vec{b} \times \vec{c}$. Then $|\vec{a} \cdot \vec{b}|$ is equal to
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Answer: (8)
Solution:
Formula: Vector product of two vectors when two vectors are parallel or perpendicular.
$\vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}, \vec{a} \cdot \vec{c}=7$
$\overrightarrow{{}a} \times \overrightarrow{{}c}-\overrightarrow{{}b} \times \overrightarrow{{}c}=\overrightarrow{{}0}$,
$(\vec{a}-\vec{b}) \times \vec{c}=0 \Rightarrow(\vec{a}-\vec{b})$ is paralleled to $\vec{c}$
$\vec{a}-\vec{b}=\mu \vec{c}$, where $\mu$ is a scalar
$-2 \hat{i}+7 \hat{j}+2 \lambda \hat{k}=\mu \cdot \overrightarrow{{}c}$
Now $\vec{a} \cdot \vec{c}=7$ gives $2 \lambda^{2}+12=7 \mu$
And $\vec{b} \cdot \vec{c}=-\frac{43}{2}$ gives $4 \lambda^{2}+82=43 \mu$
$\mu=2$ and $\lambda^{2}=1$
$|\overrightarrow{{}a} \cdot \overrightarrow{{}b}|=8$
