Vectors — Maths STD 12 Science — Question

Then $|\bar{u} \times \bar{v}|=8$ gives

$|\bar{u}||\bar{v}| \sin \theta=8$

∴ 2 × 5 × sin θ = 8

$\begin{aligned} & \therefore \sin \theta=\frac{4}{5} \\ & \cos \theta= \pm \sqrt{1-\sin ^2 \theta} \quad \ldots[\because 0 \leqslant \theta \leqslant \pi]\end{aligned}$

$\begin{aligned} & = \pm \sqrt{1-\left(\frac{4}{5}\right)^2}= \pm \sqrt{1-\frac{16}{25}} \\ & = \pm \sqrt{\frac{9}{25}}= \pm \frac{3}{5}\end{aligned}$

Now, $\bar{u} \cdot \bar{v}=|\bar{u}||\bar{v}| \cos \theta$

$\therefore \overline{u \cdot v}=2 \times 5 \times\left( \pm \frac{3}{5}\right)= \pm 6$