Question

Let $\vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}, \vec{a} \cdot \vec{c}=7,2 \vec{b} \cdot \vec{c}+43=0, \vec{a} \times \vec{c}=\vec{b} \times \vec{c}$. Then $|\vec{a} \cdot \vec{b}|$ is equal to

Answer 1

Correct Answer: 8

The correct answer is 8
$\vec{a}=\hat{i}+2\hat{j}+\lambda \hat{k},\vec{b}=3\hat{i}-5\hat{j}-\lambda \hat{k},\vec{a}\cdot \vec{c}=7$
$\vec{a}\times \vec{c}-\vec{b}\times \vec{c}=\vec{0}$
$(\vec{a}-\vec{b})\times \vec{c}=0\Rightarrow (\vec{a}-\vec{b})\text{is paralleled to}\vec{c}$
$\vec{a}-\vec{b}=\mu \vec{c}\text{, where}\mu \text{is a scalar}$
$-2\hat{i}+7\hat{j}+2\lambda \hat{k}=\mu \cdot \overrightarrow{c}$
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 ${\lambda }^2=1$
$|\vec{a}\cdot \vec{b}|=8$