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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry5 Marks
Question
Find the image of the point with position vector $3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$ in the plane $\vec{\text{r}}. (2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=4.$ Also, find the position vectors of the foot of the prependicular and the equation of the perpendicular line through $3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}.$
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Answer

Let Q be the image the point $\text{P}(3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})$ in the plane $\vec{\text{r}}.(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=4$
Since PQ passes through P and is normal to the given plane, it is parallel to the normal vector $2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$. So, the equation of PQ is
$\vec{\text{r}}=\big(3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
As Q lies on PQ, let the position vector of Q be $(3+2\lambda)\hat{\text{i}}+(1-\lambda)\hat{\text{j}}+(2+\lambda)\hat{\text{k}}.$
$=\frac{\Big[(3+2\lambda)\hat{\text{i}}+(1-\lambda)\hat{\text{j}}+(2+\lambda)\hat{\text{k}}\Big]+\Big[3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\Big]}{2}$
$=\frac{(6+2\lambda)\hat{\text{i}}+(2-\lambda)\hat{\text{j}}+(4+\lambda)\hat{\text{k}}}{2}$
$=(3+\lambda)\hat{\text{i}}+\Big(1-\frac{\lambda}{2}\Big)\hat{\text{j}}+\Big(2+\frac{\lambda}{2}\Big)\hat{\text{k}}$
Since R lies in the plane $\vec{\text{r}}.\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)=4$
$=\Big[(3+\lambda)\hat{\text{i}}+\Big(1-\frac{\lambda}{2}\Big)\hat{\text{j}}+\Big(2+\frac{\lambda}{2}\Big)\hat{\text{k}}\Big].\Big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\Big)=4$
$\Rightarrow 6+2\lambda-1+\frac{\lambda}{2}+2+\frac{\lambda}{2}=4$
$\Rightarrow 7+2\lambda+\frac{\lambda}{2}+\frac{\lambda}{2}=4$
$\Rightarrow 14+6\lambda=8$
$\Rightarrow 6\lambda=8-14$
$\Rightarrow \lambda=-1$
Putting $\lambda=-1$ in Q, we get
$\text{Q}=(3+2(-1))\hat{\text{i}}+(1-(-1))\hat{\text{j}}+(2+(-1))\hat{\text{k}}$
$=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{h}}$ or (1, 2, 1)
Therefore, by putting $\lambda=-1$ in R, we get
$\text{R}=(3+(-1))\hat{\text{i}}+\Big(1-\frac{(-1)}{2}\Big)\hat{\text{j}}+\Big(2+\frac{(-1)}{2}\Big)\hat{\text{k}}$
$=2\hat{\text{i}}+\frac{3}{2}\hat{\text{j}}+\frac{3}{2}\hat{\text{k}}$

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