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  • Find the equation of the plane through the points (2, 1, -1) and (-1, 3, 4) and perpendicular to the plane x - 2y + 4z = 10.

Find the equation of the plane through the points (2, 1, -1) and (-1, 3, 4) and perpendicular to the plane x - 2y + 4z = 10.

Answers (1)

Given, a plane passes through the points (2, 1, -1) and (-1, 3, 4) and is perpendicular to the plane x - 2y + 4z = 10.

We want to find the equation of this plane.

We know, the Cartesian equation of a plane passing through (x1, y1, z1)

with direction ratios perpendicular to a, b, c for its normal is given as:

a (x - x1) +b (y - y1) + c (z - z1) = 0

Hence,

Let us consider the equation of the plane passing through (2, 1, -1) to be

a(x – 2) + b(y – 1) + c(z – (-1)) = 0

⇒ a(x – 2) + b(y – 1) + c(z + 1) = 0 …(i)

Since it also passes through point (-1, 3, 4) we just replace x, y, z by -1, 3, and 4 respectively.

⇒ a(-1 – 2) + b(3 – 1) + c(4 + 1) = 0

⇒ -3a + 2b + 5c = 0 …(ii)

Since a, b, and c are direction ratios and this plane is perpendicular to the plane x - 2y + 4z = 10, we just replace x, y, and z with a, b, and c respectively (neglecting 10) and we equate this to 0.

=> a - 2b + 4c = 0 …(iii)

To solve two equations x1a + y1b + z1c = 0 and x2a + y2b + z2c = 0, we use the formula

\frac{a}{\begin{vmatrix} y_{1} &z_{1} \\ y_{2}&z_{2} \end{vmatrix}}=\frac{b}{\begin{vmatrix} z_{1} &x_{1} \\ z_{2}&x_{2} \end{vmatrix}}=\frac{c}{\begin{vmatrix} x_{1} &y_{1} \\ x_{2}&y_{2} \end{vmatrix}}

Similarly, to solve for equations (ii) and (iii):

\frac{a}{\begin{vmatrix}2 &5 \\ -2&4 \end{vmatrix}}=\frac{b}{\begin{vmatrix} 5 &-3 \\ 4&1 \end{vmatrix}}=\frac{c}{\begin{vmatrix}-3 & 2 \\ 1&-2 \end{vmatrix}}

\Rightarrow \frac{a}{\left ( 2 \times 4 \right )-\left ( 5 \times -2 \right )}=\frac{b}{\left ( 5 \times 1 \right )-\left ( -3 \times 4 \right )}=\frac{c}{\left ( -3 \times -2 \right )-\left ( 2 \times 1 \right )}

\Rightarrow \frac{a}{8+10}=\frac{b}{5+12}=\frac{c}{6-2}

\Rightarrow \frac{a}{18}=\frac{b}{17}=\frac{c}{4}=\lambda

\Rightarrow \frac{a}{18}=\lambda, \frac{b}{17}=\lambda, \frac{c}{4}=\lambda

That is,

\Rightarrow \frac{a}{18}=\lambda\\ \Rightarrow a=18 \lambda\\ \\ \Rightarrow \frac{b}{17}=\lambda\\ \Rightarrow b=17 \lambda\\ \\ \Rightarrow \frac{c}{4}=\lambda\\ \Rightarrow c=4 \lambda\\ \\

Substituting these values of a, b, and c in equation (i), we get

a(x - 2) + b(y - 1) + c(z + 1) = 0\\ \Rightarrow 18\lambda(x - 2) + 17\lambda(y - 1) + 4\lambda(z + 1) = 0\\ \Rightarrow \lambda[18(x - 2) + 17(y - 1) + 4(z + 1)] = 0\\ \Rightarrow 18(x - 2) + 17(y - 1) + 4(z + 1) = 0\\ \Rightarrow 18x - 36 + 17y - 17 + 4z + 4 = 0\\ \Rightarrow 18x + 17y + 4z - 36 - 17 + 4 = 0\\ \Rightarrow 18x + 17y + 4z - 49 = 0\\ \Rightarrow 18x + 17y + 4z = 49\\

Therefore, the required equation of the plane is 18x + 17y + 4z = 49.

 

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