Question 14 Marks
Show that the lines $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and $\frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}$ are coplanar. Find the equation of the plane containing them.
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Let the p.v. of points A(1, 0, 1), B(1, –1, 1) and C(4, –3, 2) be
$\vec{a}=\bar{i}+\bar{k}, \bar{b}=\bar{i}-\bar{j}+\bar{k}$ and $\bar{c}=4 \bar{i}-3 \bar{j}+2 \bar{k}$
$\bar{b}-\bar{a}=-\bar{j}, \bar{c}-\bar{a}=3 \bar{i}-3 \bar{j}+\bar{k}$
$(\bar{b}-\bar{a}) \times(\bar{c}-\bar{a})=\left|\begin{array}{ccc}\bar{i} & \bar{j} & \bar{k} \\ 0 & -1 & 0 \\ 3 & -3 & 1\end{array}\right|=-\bar{i}+3 \bar{k}$
Equation of plane through A, B, C in vector form is
$(\bar{r}-\bar{a}) \cdot[(\bar{b}-\bar{a}) \times(\bar{c}-\bar{a})]=0$
$(\bar{r}-\bar{a}) \cdot(-\bar{i}+3 \bar{k})=0$
$\bar{r} \cdot(-\bar{i}+3 \bar{k})=(\bar{i}+\bar{k}) \cdot(-\bar{i}+3 \bar{k})=-1+3=2$
$\therefore \bar{r} \cdot(-\bar{i}+3 \bar{k})=2$
The vector equation of the plane passing through the points
$A(\bar{a}), B(\bar{b})$ and $C(\bar{c})$
$\bar{r} \cdot(\overline{A B} \times \overline{A C})=\bar{a}(\overline{A B} \times \overline{A C})$......(1)
Let $\bar{a}=\hat{i}+\hat{j}-2 \widehat{k}, \bar{b}=\hat{i}+2 \hat{j}+\widehat{k}, \bar{c}=2 \hat{i}-\hat{j}+\widehat{k}$
$\therefore \overline{A B}=\bar{b}-\bar{a}=(\hat{i}+2 \hat{j}+\widehat{k})-(\hat{i}+\hat{j}-2 \widehat{k})=\hat{j}+3 \widehat{k}$
and $\overline{A C}=\bar{c}-\bar{a}=(2 \hat{i}-\hat{j}+\widehat{k})-(\hat{i}-\hat{j}-2 \widehat{k})=\hat{i}-2 \hat{j}+3 \widehat{k}$
$\therefore \overline{A B} \times \overline{A C}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 3 \\ 1 & -2 & 3\end{array}\right|$
$=(3+6) \hat{i}-(0-3) \hat{j}+(0-1) \widehat{k}$
$=9 \hat{i}+3 \hat{j}-\widehat{k}$
$\bar{a} \cdot(\overline{A B} \times \overline{A C})=(\hat{i}+\hat{j}-2 \widehat{k}) \cdot(9 \hat{i}+3 \hat{j}-\widehat{k})$
$=1(9)+1(3)+(-2)(-1)$
$=9+3+2=14$
from (1), the vector equation of the required plane is
$\bar{r} \cdot(9 \hat{i}+3 \hat{j}-\widehat{k})=14$
$(x \hat{i}+y \hat{j}+z \widehat{k})(9 \hat{i}+3 \hat{j}-\widehat{k})=14$
∴ the cartesian equation of the plane is
$9 x+3 y-z=14$
let
$\overline{A B}=\bar{b}-\bar{a}=(\hat{i}+2 \hat{j}+\widehat{k})-(\hat{i}+\hat{j}-2 \widehat{k})=\hat{j}+3 \widehat{k}$
$\overline{A C}=\bar{c}-\bar{a}=(2 \hat{i}-\hat{j}+\widehat{k})-(\hat{i}+\hat{j}-2 \widehat{k})=\hat{i}-2 \hat{j}+3 \widehat{k}$
$\overline{A B} \times \overline{A C}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \widehat{k} \\ 0 & 1 & 3 \\ 1 & -2 & 3\end{array}\right|$
$=\hat{i}(3+6)-\hat{j}(0-3)+\widehat{k}(0-1)$
$=9 \hat{i}+3 \hat{j}-\widehat{k}$
Then the equation of required plane is,
$\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}$
$\bar{r} \cdot(9 \hat{i}+3 \hat{j}-\widehat{k})=(\hat{i}+\hat{j}-2 \widehat{k}) \cdot(9 \hat{i}+3 \hat{j}-\widehat{k})$
$\bar{r} \cdot(9 \hat{i}+3 \hat{j}-\widehat{k})=9+3+2$
$\bar{r} \cdot(9 \hat{i}+3 \hat{j}-\widehat{k})=14$
The cartesian equation of the plane is given by,
$(x \hat{i}+y \hat{j}+z \widehat{k}) \cdot(9 \hat{i}+3 \hat{j}-\widehat{k})=14$
$9 x+3 y-z=14$
The cartesian equation of the plane is 9x + 3y - z = 14.
Given planes are
2x – y + z = 3 = 0,
4x – 3y + 5z + 9 = 0
Equation of required plane passing through their intersection is
(2x – y + z – 3) + λ(4x – 3y + 5z + 9) = 0 .....(1)
(2 + 4λ)x + (–1 – 3λ)y + (1 + 5λ)z + (–3 + 9λ) = 0
Direction ratios of the normal to the above plane are 2 + 4λ, –1 – 3λ and 1 + 5λ
Plane is parallel to the line $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z-3}{5}$
Direction ratios of line are 2, 4, 5
Given that required plane is parallel to given line.
∴ Normal of the plane is perpendicular to the given line
2(2 + 4λ) + 4(–1 – 3λ) + 5(1 + 5λ) = 0
4 + 8λ – 4 – 12λ + 5 + 25λ = 0
21λ + 5 = 0
21λ = – 5
$\therefore \lambda=-\frac{5}{21}$
Substituting λ in (1)
∴ Equation of plane is
$(2 x-y+z-3)-\frac{5}{21}(4 x-3 y+5 z+9)=0$
42x – 21y + 21z – 63 – 20x + 15y – 25z – 45 = 0
22x – 6y – 4z – 108 = 0
11x – 3y – 2z – 54 = 0
The equation of plane passing through the intersection of the planes 3x + 2y – z + 1 = 0 and x + y + z – 2 = 0 is
($3 x+2 y-z+1)+\lambda(x+y+z-2)=0$.........(1)
It passes through the point (2, 2, 1)
$(6+4-1+1)+\lambda(2+2+1-2)=0$
$10+3 \lambda=0$
$\lambda=-\frac{10}{3}$
Now,
$(3 x+2 y-z+1)+\left(-\frac{10}{3}\right)(x+y+z-2)=0 \quad \ldots \ldots . .[$ from(1)]
$9 x+6 y-3 z+3-10 x-10 y-10 z+20=0$
$-x-4 y-13 z+23=0$
The equation of plane is $x+4 y+13 z=23$