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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry5 Marks
Question
Find the vector equation of the plane passing through three points with position vectors $\hat{\text{i}} + \hat{\text{j}} - 2\hat{\text{k}} , 2\hat{\text{i}} - \hat{\text{j}} + \hat{\text{k}}\text{ and } \hat{\text{i}} + 2\hat{\text{j}} + \hat{\text{k}}.$Also find the coordinates of the point of intersection of this plane and the line $\overrightarrow{\text{r}} = 3\hat{\text{i}} - \hat{\text{j}} - \hat{\text{k}} + \lambda(2\hat{\text{i}}- 2 \hat{\text{j}} + \hat{\text{k}}).$
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Answer

The equation of plane passing through three points $\hat{\text{i}} + \hat{\text{j}} - 2 \hat{\text{k}} , 2\hat{\text{i}} - \hat{\text{j}} + \hat{\text{k}}\text{ and } \hat{\text{i}} +2\hat{\text{j}} + \hat{\text{k}}$
i.e., (1, 1, –2), (2, – 1, 1) and (1, 2, 1) is
$\begin{bmatrix} \text{x} - 1 & \text{y} - 1 &\text{z} +2 \$0.3em] 2- 1 & - 1 - 1 &1 + 2 \$0.3em] 1 - 1 &2 - 1 & 1 + 2 \end{bmatrix} = 0 \Rightarrow\begin{bmatrix} \text{x} - 1 & \text{y} - 1 &\text{z} +2 \$0.3em] 1 &-2 &3 \$0.3em] 0 &1 & 3 \end{bmatrix} = 0 $
$\Rightarrow$ (x – 1) ( – 6 – 3) – (y – 1) (3 – 0) +(z + 2) (1 + 0) = 0
$\Rightarrow$–9x + 9 – 3y + 3 +z + 2 = 0
$\Rightarrow$9x + 3y – z =14 - - - - (i)
Its vector form is,
$\overrightarrow{\text{r}}.(9\hat{\text{j}} + 3 \hat{\text{j}} - \hat{\text{k}} ) = 14 $
The given line is
$\overrightarrow{\text{r}}.(3\hat{\text{i}} -\hat{\text{j}} - \hat{\text{k}} ) + \lambda(2\hat{\text{i}} - 2\hat{\text{j}} +\hat{\text{k}})$
Its cartesian form is
$\frac{\text{x} - 3}{2} = \frac{\text{y} + 1 }{-2} = \frac{\text{z} + 1 }{1}$ - - - - -(ii)
Let the line (ii) intersect plane (i) at ($\alpha,\beta, \gamma$)
$\because(\alpha , \beta ,\gamma)\text{ lie on (ii) } $
$\frac{\alpha - 3}{2} = \frac{\beta + 1 }{-2} = \frac{\gamma + 1 }{1} = \lambda\text{ (say)}$
$\Rightarrow \alpha= 2\lambda + 3; \beta = – 2\lambda – 1; \gamma = \lambda – 1$
Also, point ($\alpha,\beta,\gamma$) lie on plane (i)
$\Rightarrow9 \alpha + 3\beta - \gamma= 14 $
$\Rightarrow9 ( 2 \lambda + 3 ) + 3( - 2 \lambda - 1 ) - (\lambda - 1) = 14 $
$\Rightarrow18\lambda + 27 - 6\lambda - 3 - \lambda + 1 = 14 \Rightarrow11 \lambda + 25 = 14 $
$\Rightarrow 11 \lambda = 14 - 25 \Rightarrow11 \lambda = -1 1$
$\Rightarrow\lambda = - 1 $
Therefore point of intersection º (1, 1, – 2).

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