AB =3 i - j + k and CD =-3 i +2 j +4 k are two vectors. The position vectors of the points A and C are 6 i +7 j +4 k and -9 j +2 k , respectively. Find the position vector of a point P on the line AB and a point Q on the line CD such that PQ is perpendicular to AB and CD both.

We have AB =3 i - j + k and CD =-3 i +2 j +4 k Also, the position vectors of A and C are 6 i +7 j +4 k and -9 j +2 k , respectively. Since, PQ is perpendicular to both AB and 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 CD is given by r =(6 i +7 j +4 k )+ (3 i - j - k ) ....(i) and the line throught c and parallel to the vector CD is given by r =-9 j +2 k + (-3 i +2 j +4 k ) ......(ii) Let P(6+3 ,7- ,4+ ) is any point on the first line and Q be any point on second line is given by (-3 ,-9+2 ,2+4 ). PQ =(-3 -6-3 ) i -(2 + -16) j +(4 - -2) k If PQ is perpendiculas to the first line, then 3(-3 -6-3 )-(2 + -16)+(4 - -2)=0 -7 -11 -4=0 .....(iii) If PQ is perpendiculas to the second line, then -3(-3 -6-3 )+2(2 + -16)+4(4 - -2)=0 29 +7 -22=0 On solving Eqs. (iii) and (iv), we get =1 and =-1 OP =3 i +8 j +3 k [from (i)] and OP =-3 i -7 j +6 k [from (ii)]

AB =3 i - j + k and CD =-3 i +2 j +4 k are two vectors. The posit…

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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)]