Question

obtain the equation of the plane normal to the yz-plane and passing through the points (1,-2,4) and (3,-4,5)​

Answer 1

Let there exist a point $(x,\ y,\ z)$ on the plane passing through the points (1, -2, 4) and (3, -4, 5).

The following vectors can be obtained by joining any two among the three points.

• $\mathbf{b_1}=\left<2,\ -2,\ 1\right>$

• $\mathbf{b_2}=\left<x-1,\ y+2,\ z-4\right>$

• $\mathbf{b_3}=\left<x-3,\ y+4,\ z-5\right>$

We take cross product of the first two vectors, which is perpendicular to the plane.

$\longrightarrow\mathbf{b_1\times b_2}=\left|\begin{array}{ccc}\mathbf{\hat i}&\mathbf{\hat j}&\mathbf{\hat k}\\2&-2&1\\x-1&y+2&z-4\end{array}\right|$

$\small\longrightarrow\mathbf{b_1\times b_2}=\left<-2(z-4)-(y+2),\ (x-1)-2(z-4),\ 2(y+2)+2(x-1)\right>$

$\small\longrightarrow\mathbf{b_1\times b_2}=\left<-y-2z+6,\ x-2z+7,\ 2x+2y+2\right>$

This vector is perpendicular to the plane.

Since the plane is normal to YZ plane, this vector should be parallel to the YZ plane, which means the vector has no component along X direction, i.e.,

the x-component of this vector is zero.

$\longrightarrow -y-2z+6=0$

or,

$\longrightarrow\underline{\underline{y+2z-6=0}}$

This is the equation of the required plane.

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

• Let us assume that the direction ratios of the plane be (a, b, c).

Now,

• Equation of plane pasing through the points (1, - 2, 4) and having direction ratios (a, b, c) is given by

$\bf \: a(x - 1) + b(y + 2) + c(z - 4) = 0 - - (1)$

Now,

• Plane (1) passes through the point (3, - 4, 5),

So,

• we get

$\bf \: a(3 - 1) + b ( - 4 + 2) + c(5 - 4) = 0$

$\bf :\implies\:2a - 2b + c = 0 - - (2)$

Further,

• Plane (1) is || to yz - plane,

therefore,

• it must be normal to x - axis whose dr’s are (1, 0, 0).

So,

$\rm :\implies\:a.1 + b.0 + c.0 = 0$

$\bf :\implies\:a \: = \: 0 - - (3)$

This implies,

Equation (2) can be rewritten as

$\rm :\implies\:2 \times 0 - 2b + c = 0$

$\bf :\implies\:2b \: = \: c \: - - - (4)$

Now,

Substituting the values from (3) and (4) in equation (1),

$\rm :\implies\:\tt \: 0 \times (x - 1) + b(y + 2) + 2b(z - 4) = 0$

$\rm :\implies\:b(y + 2 + 2z - 8) = 0$

$\rm :\implies\:y + 2z - 6 = 0 \: is \: the \: required \: plane$