Question

Find the angles between the following pairs of lines:
(i) $\overrightarrow r= 2\hat i-5\hat j+\hat k+ \lambda (3\hat i-2\hat j+6\hat k)$and

$\overrightarrow r=7\hat i-6\hat k+\mu(\hat i+2\hat j+2\hat k)$

(ii) $\overrightarrow r=3\hat i+\hat j-2\hat k+\lambda(\hat i-\hat j-2\hat k)$ and

$\overrightarrow r = 2\hat i-\hat j-56\hat k+\mu(3\hat i-5\hat j-4\hat k)$

Answer 1

(i) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by cosQ=$\begin{vmatrix}\frac{\overrightarrow b. \overrightarrow b}{\| \overrightarrow b \mid \mid \overrightarrow b. \|} \end{vmatrix}$

The given lines are parallel to the vectors, $\overrightarrow b_1=3\hat i+2\hat j+6\hat k\, and \overrightarrow b_2=\hat i+2\hat j+2\hat k$, respectively,

∴|$\overrightarrow b_1$|=$\sqrt {3^2+2^2+6^2}$ =7

|$\overrightarrow b_2$|=$\sqrt{(1)^2+(2)^2+(2)^2}=3$

$\overrightarrow b_1.\overrightarrow b_2$

=$(3\hat i+2\hat j+6\hat k\,)(\hat i+2\hat j+2\hat k)$

=31+22+62
=3+4+12
=19
$\Rightarrow$ cos Q=$\frac{19}{73}$
$\Rightarrow$ Q= \cos^{-1}$$\bigg(\frac{19}{21}\bigg)

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(ii) The given lines are parallel to the vectors $\overrightarrow b_1=\hat i-\hat j-2\hat k$ and $\overrightarrow b_2=3\hat i-5\hat j-4\hat k$, respectively,

∴$\mid \overrightarrow b_1\mid= \sqrt {(1)^2+(-1)^2+(-2)^2}=\sqrt 6$

$\mid\overrightarrow b_1 \mid= \sqrt {(3)^2+(-5)^2+(-4)^2}=\sqrt {50}=5\sqrt 2$

$\overrightarrow b_1.\overrightarrow b_2$

=$(\hat i-\hat j-2\hat k )$.$(3\hat i-5\hat j-4\hat k )$

=1.3-1(-5)-2(-4)

=3+5+8

=16
cos Q=$\begin{vmatrix}\frac{\overrightarrow b. \overrightarrow b}{\| \overrightarrow b \mid \mid \overrightarrow b. \|} \end{vmatrix}$

cos Q=$\frac{16}{\sqrt16.5\sqrt2}$

cos Q=$\frac{16}{\sqrt 2.\sqrt3.5\sqrt2}$

cos Q=$\frac{16}{10\sqrt3}$

cos Q=$\frac{8}{5\sqrt3}$

Q=cos-1$\frac{8}{5\sqrt3}$