Question

Match the following
$\begin{array}{cccc}& Column-I& & Column-II\\ \left(A\right)& Theshortestdistancebetweenthelines& \left(P\right)& 2\\ & \frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}and\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}is& & \\ \left(B\right)& Thedistanceofthepointofintersectionoftheline& \left(Q\right)& 3\sqrt{30}\\ & \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}andtheplane& & \\ & x-y+z=5fromthepoint(-1,-5,-10)is& & \\ \left(C\right)& Thelengthoftheperpendiculardrawnfromthe& \left(R\right)& 13\\ & point(5,4,-1)ontheline\frac{x-1}{2}=\frac{y}{9}=\frac{z}{5}is& & \\ \left(D\right)& ThedistancebetweenthepointsPandQisdand& \left(S\right)& \sqrt{\frac{2109}{110}}\\ & thelenghtoftheprojectionofPQontheco-ordinate& & \\ & planesare{d}_1,d_2,d_{3}then{d}_1^2+d_2^2+d_3^2=k{d}^{2}whenkis& & \end{array}$

• A. $A\to R,B\to q,C\to s,D\to P$
• B. $A\to q,B\to r,C\to s,D\to P$
• C. $A\to q,B\to s,C\to r,D\to P$
• D. $A\to p,B\to s,C\to r,D\to q$

Answer 1

The correct option is B $A\to q,B\to r,C\to s,D\to P$
A. $\frac{\overline{r}=\overline{a}+t.\overline{b}}{\overline{r}=\overline{c}+s\overline{d}}$ shortest distance between the lines = $=\frac{|[\overline{c}-\overline{a}\overline{b}\overline{d}]|}{|\overline{b}\times \overline{d}|}$
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)
$\left(\right(2k-4)\overline{i}+(9k-4)+(5k+1\left)\overline{k}\right).(2\overline{i}+9\overline{j}+5\overline{k})=0$ find k.
d. projection of PQ on xy-plane = $\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$
projection of PQ on yz-plane = $\sqrt{(y_1-y_2{)}^2+(z_1-z_2{)}^2}$
projection of PQ on zx-plane = $\sqrt{(x_1-x_2{)}^2+(z_1-z_2{)}^2}$