Question

Match the following :

$\begin{array}{cccc}& coloumn1& & coloumn2\\ \left(A\right)& Theshortestdistancebetweenthelines& \left(P\right)& 13\\ & \frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}and\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}& & \\ \left(B\right)& Thedistanceofthepointofintersectionoftheline& \left(Q\right)& \sqrt{\frac{2109}{110}}\\ & \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}andtheplanex-y+z=5fromthepoint& & \\ & (-1,-5,-10)is& & \\ \left(C\right)& Thelengthoftheperpendiculardrawnfromthepoint(5,4,-1)& \left(R\right)& 2\\ & ontheline\frac{x-1}{2}=\frac{y}{9}=\frac{z}{5}is& & \\ \left(D\right)& ThedistancebetweenthepointsPandQisdandthelengthof& \left(S\right)& 3\sqrt{30}\\ & theprojectionofPQontheco-ordinateplanesare{d}_1,d_2,d_{3}then& & \\ & {d_1}^2+{d_2}^2+k{d}^{2}whenkis& & \end{array}$

• A. AP, BR, CS, DQ
• B. AQ, BS, CP, DR
• C. AS, BP, CQ, DR
• D. AR, BQ, CR, DP

Answer 1

The correct option is C

AS, BP, CQ, DR

A$\to$S, B$\to$P, C$\to$Q, D$\to$R

$$\begin{array}{ccc}\left(A\right)\overline{r}=\overline{a}+t\overline{b}& shortestdistancebetweentheselines& =\underset{–}{\left|\right[\overline{c}-\overline{a}\overline{b}\overline{d}\left]\right|}\\ \overline{r}=\overline{c}+s\overline{d}& & |\overline{b}\times \overline{d}|\end{array}$$

(B) (2,-1,2) satisfies the plane

find distance between (2,-1,2) and (-1,-5,-10)

(C) Find d.r's of A(5,4,-1) and B (1+2k, 9k, 5k)

((2k-4)$\overline{i}$ + (9k -4) + (5k + 1) $\overline{k})$.(2$\overline{i}$+9$\overline{j}$+5$\overline{k}$) =0 find k.

(d) projection of PQ on xy- plane= $\sqrt{{(x_1-x_2)}^2+(y_1-y_2}{)}^2$

projection of PQ on xy- plane = $\sqrt{{(y_1-y_2)}^2+(z_1-z_2}{)}^2$

projection of PQ on xy- plane = $\sqrt{{(x_1-x_2)}^2+(z_1-z_2}{)}^2$