Let a line L be perpendicular to both the lines L_1:(x+1)/(3)=(y+3)/(5)=(z+5)/(7) and L_2:…

Question

Let a line L be perpendicular to both the lines ${L}_{1}:\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and ${L}_{2}:\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-6}{7}$.

If $	heta $ is the acute angle between the lines L and ${L}_{3}:\frac{x-\frac{8}{7}}{2}=\frac{y-\frac{4}{7}}{1}=\frac{z}{2}$, then $	an 	heta $ is equal to:

Accepted Answer

For a line $L$ to be perpendicular to both lines ${L}_{1}$ and ${L}_{2}$, its direction ratios must be perpendicular to the direction ratios of both ${L}_{1}$ and ${L}_{2}$.

So first let us find the direction vectors.

For ${L}_{1}$:

$\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$

Direction ratios of ${L}_{1}$ are

$\vec{{d}_{1}}=(3,5,7)$

For ${L}_{2}$:

$\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-6}{7}$

Direction ratios of ${L}_{2}$ are

$\vec{{d}_{2}}=(1,4,7)$

Since line $L$ is perpendicular to both, its direction vector will be parallel to

$\vec{d}=\vec{{d}_{1}}	imes \vec{{d}_{2}}$

Now,

$\vec{{d}_{1}}	imes \vec{{d}_{2}}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 3 \quad  5 \quad  7 \ 1 \quad  4 \quad  7 \end{vmatrix}$

$=\hat{i}(35-28)-\hat{j}(21-7)+\hat{k}(12-5)$

$=7\hat{i}-14\hat{j}+7\hat{k}$

$=7(1,-2,1)$

So direction ratios of line $L$ can be taken as

$(1,-2,1)$

Now for ${L}_{3}$:

$\frac{x-\frac{8}{7}}{2}=\frac{y-\frac{4}{7}}{1}=\frac{z}{2}$

Direction ratios of ${L}_{3}$ are

$(2,1,2)$

Let $	heta $ be the acute angle between $L$ and ${L}_{3}$.

Using the formula

$	an 	heta =\frac{|\vec{a}	imes \vec{b}|}{|\vec{a}\cdot \vec{b}|}$

where

$\vec{a}=(1,-2,1),\, \vec{b}=(2,1,2)$

First find dot product:

$\vec{a}\cdot \vec{b}=1\cdot 2+(-2)\cdot 1+1\cdot 2$

$=2-2+2=2$

Now find cross product:

$\vec{a}	imes \vec{b}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 1 \quad  -2 \quad  1 \ 2 \quad  1 \quad  2 \end{vmatrix}$

$=\hat{i}((-2)(2)-1\cdot 1)-\hat{j}(1\cdot 2-1\cdot 2)+\hat{k}(1\cdot 1-(-2)\cdot 2)$

$=\hat{i}(-4-1)-\hat{j}(2-2)+\hat{k}(1+4)$

$=-5\hat{i}+0\hat{j}+5\hat{k}$

So,

$|\vec{a}	imes \vec{b}|=\sqrt{(-5{)}^{2}+{0}^{2}+{5}^{2}}=\sqrt{50}=5\sqrt{2}$

Hence,

$	an 	heta =\frac{5\sqrt{2}}{2}$

Therefore,

$	an 	heta =\frac{5}{2}\sqrt{2}$

So the correct option is:

Solutions · 1