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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry5 Marks
Question
$\overrightarrow{\text{AB}}=3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\overrightarrow{\text{CD}}=-3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$ are two vectors. The position vectors of the points A and C are $6\hat{\text{i}}+7\hat{\text{j}}+4\hat{\text{k}}$ and $-9\hat{\text{j}}+2\hat{\text{k}},$ respectively. Find the position vector of a point P on the line AB and a point Q on the line CD such that $\overrightarrow{\text{PQ}}$ is perpendicular to $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{CD}}$ both.
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Answer

We have $\overrightarrow{\text{AB}}=3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\overrightarrow{\text{CD}}=-3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
Also, the position vectors of A and C are $6\hat{\text{i}}+7\hat{\text{j}}+4\hat{\text{k}}$ and $-9\hat{\text{j}}+2\hat{\text{k}},$ respectively.
Since, $\overrightarrow{\text{PQ}}$ is perpendicular to both $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{CD}}.$
So, P and Q will be foot of perpendicular to both the liens through A abd C.
Now, equation of the through C and parallel to the vector $\overrightarrow{\text{CD}}$ is given by
$\vec{\text{r}}=(6\hat{\text{i}}+7\hat{\text{j}}+4\hat{\text{k}})+\lambda(3\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}})\ ....(\text{i})$
and the line throught c and parallel to the vector $\overrightarrow{\text{CD}}$ is given by
$\vec{\text{r}}=-9\hat{\text{j}}+2\hat{\text{k}}+\mu(-3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}})\ ......(\text{ii})$
Let $\text{P}(6+3\lambda,7-\lambda,4+\lambda)$ is any point on the first line and Q be any point on second line is given by $(-3\mu,-9+2\mu,2+4\mu).$
$\therefore\overrightarrow{\text{PQ}}=(-3\mu-6-3\lambda)\hat{\text{i}}-(2\mu+\lambda-16)\hat{\text{j}}+(4\mu-\lambda-2)\hat{\text{k}}$
If $\overrightarrow{\text{PQ}}$ is perpendiculas to the first line, then
$3(-3\mu-6-3\lambda)-(2\mu+\lambda-16)+(4\mu-\lambda-2)=0$
$\Rightarrow-7\mu-11\lambda-4=0\ .....(\text{iii})$
If $\overrightarrow{\text{PQ}}$ is perpendiculas to the second line, then
$-3(-3\mu-6-3\lambda)+2(2\mu+\lambda-16)+4(4\mu-\lambda-2)=0$
$\Rightarrow29\mu+7\lambda-22=0$
On solving Eqs. (iii) and (iv), we get
$\mu=1$ and $\lambda=-1$
$\therefore\overrightarrow{\text{OP}}=3\hat{\text{i}}+8\hat{\text{j}}+3\hat{\text{k}}$ [from (i)]
and $\overrightarrow{\text{OP}}=-3\hat{\text{i}}-7\hat{\text{j}}+6\hat{\text{k}}$ [from (ii)]

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