Question

find the equation of the plane containing the lines r vector = i+j+lambda(i+2j-k)and r vector=i+j+nu(-i+j-2k).find the distance of this plane from the origin and also from the point (1,1,1).

Answer 1

the equation of the plane containing the lines $(\hat i+\hat j)+\lambda(\hat i+2\hat j-\hat k)$ and $(\hat i+\hat j)+\mu(\-hat i+\hat j-2\hat k)$ is $\boxed{-x+y+z=0}$

The perpendicular distance from the origin is 0

The perpendicular distance from point $(1,1,1)$ is $\frac{1}{\sqrt{3}}$

Step-by-step explanation:

The given equation of lines

$\vec r=(\hat i+\hat j)+\lambda(\hat i+2\hat j-\hat k)$

or, $\vec r=\vec a+\lambda\vec b$

And

$\vec r=(\hat i+\hat j)+\mu(\-hat i+\hat j-2\hat k)$

or, $\vec r=\vec a+\mu\vec b'$

Here,

$\vec b=\hat i+2\hat j-\hat k$

And $\vec b'=-\hat i+\hat j-2\hat k$

The equation of the plane containing the lines is given by

$\vec r.\vec n=\vec a.\vec n$

Where,

$\vec n=\vec b\times \vec b'$

$\implies \vec n=\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&-1\\-1&1&-2\end{vmatrix}$

$\implies \vec n=\hat i(-4+1)-\hat j(-2-1)+\hat k(1+2)$

$\implies \vec n=-3\hat i+3\hat j+3\hat k$

Therefore, the equation of the plane

$\vec r.(-3\hat i+3\hat j+3\hat k)=(\hat i+\hat j).(-3\hat i+3\hat j+3\hat k)$

$\implies (x\hat i+y\hat j+z\hat k).(-3\hat i+3\hat j+3\hat k)=-3+3$

$\implies -3x+3y+3z=0$

$\implies -x+y+z=0$

This is the equation of the required plane.

Perpendicular distance of this plane from point $(1,1,1)$ is given by

$d=\frac{-1+1+1}{\sqrt{(-1)^2+1^2+1^2}}$

$\implies d=\frac{1}{\sqrt{3}}$

$\implies d=1/\sqrt{3}$ units

Perpendicular distance of the plane from origin $(0,0,0$ is given by

$d=\frac{0}{\sqrt{(-1)^2+1^2+1^2}}=0$

Hope this answer is helpful.

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