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Column I Column II
P P is a point on the segment of the line joining (3, 3, 5) and (4, 6, 7) dividing it in the ratio k : 1 and also projection OP(where O is the origin) on coordinate axes are $\frac{32}{9}, \frac{42}{9}, \frac{55}{9}$ , then value of k is 1 19
Q A (3, 4, 6), B(0, 4, 3), C(a, 6, b) are the vertices of a triangle ABC such that the median through A is equally inclined to the coordinate axes, then $(a + b)$ is 2 35
R The line $\frac{x + 1}{- 1} = \frac{y - 12}{5} = \frac{z - 7}{2}$ intersects the curve $y z = k, x = 0,$ then the value of k is 3 15
S Given $| \vec{p} | = 2 \sqrt{2}, | \vec{q} | = 3,$ angle between $\vec{p} a n d \vec{q}$ is $\frac{π}{4}$ . If $\vec{a} = \vec{p} - 3 \vec{q}, \vec{b} = 5 \vec{p} + 2 \vec{q}$ form the adjacent sides of a parallelogram then the length of one of the diagonals is 4 5/4

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$P \to 2; Q \to 1; R \to 4; S \to 3$

$P \to 4; Q \to 1; R \to 2; S \to 3$

$P \to 3; Q \to 2; R \to 3; S \to 4$

$P \to 3; Q \to 2; R \to 4; S \to 1$

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Detailed Solution

(P) $\vec{O P} = \frac{32}{9} \hat{i} + \frac{42}{9} \hat{j} + \frac{55}{9} \hat{k}$
$\frac{4 k + 3}{k + 1} = \frac{32}{9}, \frac{6 k + 3}{k + 1} = \frac{42}{9}, \frac{7 k + 5}{k + 1} = \frac{55}{9}$
All these give k = $\frac{5}{4}$
(Q) mid point D of BC = $(\frac{9}{2}, 5, \frac{b + 3}{2})$
direction ratios of median AD are $3 - \frac{a}{2}, - 1, 6 - (\frac{b + 3}{2})$

Question Image
this is equally inclined to the axes
$3 - \frac{a}{2} = - 1, 6 - (\frac{b + 3}{2}) = - 1$
a = 8, b = 11
a + b = 19
(R) Solve with x = 0 yz = 35, k = 35
$\frac{y - 12}{5} = - 1 \Rightarrow y = 7, \frac{z - 7}{2} = - 1 \Rightarrow z = 5,$
(S) One diagonal is given by $\vec{a} + \vec{b} = 6 \vec{p} - \vec{q}$
(length)2 = 36 $| \vec{p} |^{2} + | \vec{q} |^{2} - 12 | \vec{p} | | \vec{q} | \cos \frac{π}{4} = 225$
length = $\sqrt{225} = 15,$ other diagonal = $4 \vec{p} + 5 \vec{q}$
${(l e n g t h)}^{2} = 16 \times 8 + 25 \times 9 + 40 \times 2 \sqrt{2} \times 3 \times \frac{1}{\sqrt{2}}$ length of other diagonal = $\sqrt{593}$
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