Find the angle between two lines, one of which has direction ratios 2, 2, 1 while the other one is obtained by joining the points (3, 1, 4) and (7, 2, 12).

Open in App

Solution

The direction ratios of the line joining the points (3, 1, 4) and (7, 2, 12) are proportional to 4, 1, 8.

Let $\vec{m_{1}} and \vec{m_{2}}$ be vectors parallel to the lines having direction ratios proportional to 2, 2, 1 and 4, 1, 8.

Now,
$\vec{b_{1}} = 2 \hat{i} + 2 \hat{j} + \hat{k} \vec{b_{2}} = 4 \hat{i} + \hat{j} + 8 \hat{k}$

If $θ$ is the angle between the given lines, then

$\cos θ = \frac{\vec{m_{1}} . \vec{m_{2}}}{|\vec{m_{1}}| |\vec{m_{2}}|} = \frac{(2 \hat{i} + 2 \hat{j} + \hat{k}) . (4 \hat{i} + \hat{j} + 8 \hat{k})}{\sqrt{2^{2} + 2^{2} + 1^{2}} \sqrt{4^{2} + 1^{2} + 8^{2}}} = \frac{8 + 2 + 8}{3 \times 9} = \frac{2}{3} \Rightarrow θ = \cos^{- 1} (\frac{2}{3})$

Suggest Corrections

Similar questions

2 : 2 : 1

(3, 1, 4)

(7, 2, 12)

(2, 2, 1)

(- 4, 1, 8)

(2, 1, 3)

(1, - 1, 2)

2, 1, - 1

(2, 1, 3)

(1, - 1, 2)

2, 1, - 1 i s \frac{2 π}{m}

m

Join BYJU'S Learning Program