Question

Find the equation of a plane passing through the points A(2,1,2) and
B(4, -2,1) and perpendicular to planer. (î - 2k) = 5. Also, find the
coordinates of the point, where the line passing through the points (3,4,1)
and (5,1,6) crosses the plane thus obtained.​

Answer 1

Therefore the required equation of plane is

6x +3y +3z -18=0

Therefore the coordinate of the intersection point is $(\frac{4}{3} ,\frac{13}{2} ,\frac{-19}{6})$

Step-by-step explanation:

Given , equation  of plane  $\vec {r}. (\hat {i}- 2 \hat k) = 5$

the direction ratio of the above equation of plane is (1,0,-2)

Let, the equation of plane which passes through the point (2,1,2) is

$a(x - 2) +b (y - 1) +c (z-2)=0$............(1)

Since the point (4,-1,1) lies on the above plane. So it will be satisfy the equation of plane.

$a(4-2)+b(-2-1)+c(1-2) = 0$

⇔$2a-3b -c=0$................(2)

Again this plane is perpendicular to the given equation of plane

So ,$l_1l_2+m_1m_2+n_1n_2=0$

⇔a×1 +b×(0) + c×(-2)=0

⇔a-2c=0...........(3)

Eliminating a, band c from equation (1),(2) and (3)

we get

$\left|\begin{array}{ccc}x-2&y-1&z-2\\2&-3&-1\\1&0&-2\end{array}\right| =0$

⇔ 6x +3y +3z -18=0

Therefore the required equation of plane is

6x +3y +3z -18=0

The equation of plane which passes through the points (3,4,1) and (5,1,6) is

$\frac{x-3}{3-5} =\frac{y-4}{4-1} =\frac{z-1}{1-6}$

$\frac{x-3}{-2} =\frac{y-4}{3} =\frac{z-1}{-5} = r (say)$

Any point on the straight line be P(-2r+3,3r+4,-5r+1)

Let P be the intersection point of the plane and the e satisfy the straight line.

So P will be satisfy the equation of plane

Then, 6(-2r+3)+3(3r+4)+3(-5r+1) - 18 = 0

⇔-18r +15 =0

⇔$r =\frac{5}{6}$

Therefore the coordinate of the intersection point is $(\frac{ -2 \times5}{6} +3,\frac{3 \times 5}{6} +4,\frac{-5\times 5}{6} +1)$ = $(\frac{4}{3} ,\frac{13}{2} ,\frac{-19}{6})$