Let a=2i-3j+4k,b=3i+4j-5k and a vector c be such that a (b+c)+b c=i+8j+13k.

Question

Let $\vec{a}=2\hat{i}-3\hat{j}+4\hat{k},\vec{b}=3\hat{i}+4\hat{j}-5\hat{k}$ and a vector $\vec{c}$ be such that $\vec{a}	imes (\vec{b}+\vec{c})+\vec{b}	imes \vec{c}=\hat{i}+8\hat{j}+13\hat{k}$. If $\vec{a}\cdot \vec{c}=13$, then $(24-\vec{b}\cdot \vec{c})$ is equal to \_\_\_\_\_\_\_.

Answer: 46

Accepted Answer

Let $\hat{i}+8\hat{j}+13\hat{k}=\vec{u}$

Given $\vec{a}	imes (\vec{b}+\vec{c})+\vec{b}	imes \vec{c}=\vec{u}$

$\begin{gathered} \Rightarrow \, \vec{a}	imes \vec{b}+\vec{a}	imes \vec{c}+\vec{b}	imes \vec{c}=\vec{u} \ (\vec{a}+\vec{b})	imes c=\vec{u}-\vec{a}	imes \vec{b} \end{gathered}$

Taking cross product with $\vec{a}$ on both sides

$\vec{a}	imes ((\vec{a}+\vec{b})	imes \vec{c})=\vec{a}	imes (\vec{u}-\vec{a}	imes \vec{b})$

$\Rightarrow \, \vec{c}\cdot ({\vec{a}}^{2}+\vec{a}\cdot \vec{b})=13(\vec{a}+\vec{b})-\vec{a}	imes \vec{u}+(\vec{a}\cdot \vec{b})\cdot \vec{a}-{\vec{a}}^{2}\vec{b}\,$ ${\because \vec{a}.\vec{c}=13}$

Putting the values, $\vec{c}=(-1,-1,3)$

$\vec{b}.\vec{c}=-22$

$\Rightarrow 24-\vec{b}\cdot \vec{c}=46$

Solutions · 1