Let a=i+2j+3k and b=i+j-k.

Question

Let $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}-\hat{k}$. If $\vec{c}$ is a vector such that $\vec{a}\cdot \vec{c}=11,\vec{b}\cdot (\vec{a}	imes \vec{c})=27$ and $\vec{b}\cdot \vec{c}=-\sqrt{3}|\vec{b}|$, then $|\vec{a}	imes \vec{c}{|}^{2}$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 285

Accepted Answer

Given,

$\begin{gathered} \vec{a}=\hat{i}+2\hat{j}+3\hat{k} \ \vec{b}=\hat{i}+\hat{j}-\hat{k} \ \vec{a}\cdot \vec{c}=11 \ \vec{b}\cdot (\vec{a}	imes \vec{c})=27 \ \vec{b}\cdot \vec{c}=-\sqrt{3}|\vec{b}| \ (\vec{b}	imes \vec{a})\cdot \vec{c}=27 \end{gathered}$

$\vec{c}={c}_{1}\hat{i}+{c}_{2}\hat{j}+{c}_{3}\hat{k}$

As $\vec{a}\cdot \vec{c}=11$

$	herefore $ ${c}_{1}+2{c}_{2}+3{c}_{3}=11$ ......(i)

Also, $\vec{b}\cdot \vec{c}=-\sqrt{3}|\vec{b}|$

$\begin{gathered} 	herefore {c}_{1}+{c}_{2}-{c}_{3}=-\sqrt{3}\sqrt{3} \ \Rightarrow {c}_{1}+{c}_{2}-{c}_{3}=-3......(ii) \end{gathered}$

Also, $\vec{b}\cdot (\vec{a}	imes \vec{c})=27$

$	herefore $ $5{c}_{1}-4{c}_{2}+{c}_{3}=27$ ...........(iii)

From (i), (ii) \& (iii)

$\vec{c}=3\hat{i}-2\hat{j}+4\hat{k}$

$\begin{gathered} |\vec{a}	imes \vec{c}{|}^{2}={|\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ 1 \quad  2 \quad  3 \ 3 \quad  -2 \quad  +4\end{array}|}^{2} \ =|14\hat{i}+5\hat{j}-8\hat{k}{|}^{2} \ ={14}^{2}+{5}^{2}+{8}^{2}=285 \end{gathered}$

Solutions · 1