If $ heta$ is the angle between $\vec{f}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{g}=2 \hat{i}-3 \hat{j}+a \hat{k}$ and $\sin heta=\sqrt{\frac{24}{28}}$ then $7 a^2+24 a=$

Question

If $	heta$ is the angle between $\vec{f}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{g}=2 \hat{i}-3 \hat{j}+a \hat{k}$ and $\sin 	heta=\sqrt{\frac{24}{28}}$ then $7 a^2+24 a=$

Aicharya · Accepted Answer

$\begin{aligned} & \vec{f}=\hat{i}+2 \hat{j}-3 \hat{k}, \vec{g}=2 \hat{i}-3 \hat{j}+a \hat{k} \\ & |\vec{f} \times \vec{g}|=|\vec{f}||\vec{g}| \sin \theta ....(i) \end{aligned}$
Now, $\vec{f} \times \vec{g}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & -3 & a\end{array}\right|=(2 a-9) \hat{i}-(a+6) \hat{j}-7 \hat{k}$ $\begin{aligned} & |\vec{f} \times \vec{g}|=\sqrt{(2 a-9)^2+(a+6)^2+49} \\ & =\sqrt{5 a^2-24 a+166} \end{aligned}$
From (i) we get, $\begin{aligned} & 5 a^2-24 a+166=14 \times\left(12+a^2\right) \times \frac{24}{28} \\ & \Rightarrow 7 a^2+24 a=10 \end{aligned}$

If $\theta$ is the angle between $\vec{f}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{g}=2 \hat{i}-3 \hat{j}+a…

Solution