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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry5 Marks
Question
If the lines $\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{-2\text{k}}=\frac{\text{z}-3}{2}$ and $\frac{\text{x}-1}{\text{k}}=\frac{\text{y}-2}{1}=\frac{\text{z}-3}{5}$ are perpendicular, find the value of $k$ and, hence find the equation of the plane containing these lines.
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Answer

We know that the lines $\frac{\text{x}-\text{x}_1}{\text{l}_1}=\frac{\text{y}-\text{y}_1}{\text{m}_1}=\frac{\text{z}-\text{z}_1}{\text{n}_1}$ and $\frac{\text{x}-\text{x}_2}{\text{l}_2}=\frac{\text{y}-\text{y}_2}{\text{m}_2}=\frac{\text{z}-\text{z}_2}{\text{n}_2}$ are perpendicular if
$l_1l_2 + m_1m_2 + n_1n_2= 0$
Here,
$l_1 = -3, m_1 = -2k, n_1 = 2, l_2 = k, m_2 = 1, n_2 = 5$
It is given that given are perpendicular.
$\Rightarrow l_1l_2 + m_1m_2 + n_1n_2= 0$
$\Rightarrow (-3)(k) + (-2k)(1) + (2)(5) = 0$
$\Rightarrow -3k - 2k + 10 = 0$
$\Rightarrow -5k = -10$
$⇒ k = 2$
Substituting this value in the given equation of the lines, we get
$\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{-4}=\frac{\text{z}-3}{2}\ ...(\text{i})$
$\frac{\text{x}-1}{2}=\frac{\text{y}-2}{1}=\frac{\text{z}-3}{5}\ ...(\text{ii})$
Finding the equation of the plane
Let the direction ratios of the required plane be proporional to $a, b, c.$
We know from $(i)$ and $(ii)$ that lines $(i)$ and $(ii)$ pass through the point $(1, 2, 3) $ and the direction ratios of $(i)$ and $(ii)$ are proportional to $-3, -4, 2$ and $2, 1, 5$ respectively.
Since the plane contains the lines $(i)$ and $(ii),$ the plane must pass through the point $(1, 2, 3)$ and it must be parallel to the line.
So, the equation of the plane is
$a(x - 1) + b(y - 2) + c(z - 3) = 0 ....(iii)$
$-3a - 4b + 2c = 0 ....(iv)$
$2a + b + 5c = 0 ....(v)$
Solving $(i), (ii)$ and $(iii)$ we get
$\begin{vmatrix}\text{x}-1&\text{y}-2&\text{z}-3\\-3&-4&2\\2&1&5 \end{vmatrix}=0$
$⇒ -22(x - 1) + 19(y - 2) + 5(z - 3) = 0$
$⇒ -22x + 19y + 5z = 31$

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