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

$ \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 $\lambda^2=1$
$|\vec{ a } \cdot \vec{ b }|=8$