10. vector algebra — ગણિત ધોરણ 12 સાયન્સ — Question

Given that, $\vec{a}=8 \vec{b}$ and $\vec{c}=-7 \vec{b}$

Let

$b=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}$

Then,

$\vec{a}=8\left(b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\right)$

And

$\vec{c}=-7\left(b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\right)$

Consider that angle between $\vec{a}$ and $\vec{c}$ is $x$.

Thus, $\cos x=\frac{a \cdot c}{|a||c|}$

$\cos x=\frac{-56\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)}{7 \times 8 \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}} \cdot \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}}$

$\cos x=-1$

Therefore,

$x=\pi$

hence angle between $\vec{a}$ and $\vec{c}$ is $\pi$