Column – IColumn – III)Image of the point P(3,5,7) in the plane 2x+y+z=0P)(1,2,8)II)Foot of the perpendicular from (7,14,5) on to the plane 2x+4y−z=2Q)(−3,5,6)III) Let A→=2i→+k→,B→=i→+j→+k→, C→=4i→−3j→+7k→ . If R→×B→=C→×B→ and R→.A→=0 then coordinates of R→ are R)(−1,−8,2)IV)If a→=−i→+j→+k→,b→=2i→+k→ then coordinates of a vector 2x→ which is coplanar with a→ and b→ and perpendicular to vector b→ and a→.x→=7S)(−9,−1,1)

I→S,II→P,III→R,IV→Q(a) Let image be (α,β,γ) α−32=β−51=γ−71=−2(18)6 α=−9,β−1,γ=1 (b) Let foot be (α,β,γ) α−72=β−144=γ−5−1=−(65−2)21 α−72=−3⇒α=1, β−14=−12⇒β=2, γ−5=3⇒γ=8 (c) R→×B→=C→×B→ R→−C→=λB→ R→=4i→−3j→+7k→+λ(i→+j→+k→) R→=(4+λ)i→+(λ−3)j→+(7+λ)k→ R→.A→=0⇒λ=−5 ∴R→=(−1,−8,2) (d) x→=λ(b→×(a→×b→))=λ((b→.b→)a→−(b→.a→)b→)=λ(5(−i→+j→+k→)+1(2i→+k→)) x→=λ(−3i→+5j→+6k→) Now a→.x→=7⇒3λ+5λ+6λ=7⇒λ=12 ∴2x→=−3i→+5j→+6k→

Column – IColumn – III)Image of the point P(3,5,7) in the plane 2x+y+z=0P)(1,2,8)II)Foot of the perpendicular from (7,14,5) on to the plane 2x+4y−z=2Q)(−3,5,6)III) Let A→=2i→+k→,B→=i→+j→+k→, C→=4i→−3j→+7k→ . If R→×B→=C→×B→ and R→.A→=0 then coordinates of R→ are R)(−1,−8,2)IV)If a→=−i→+j→+k→,b→=2i→+k→ then coordinates of a vector 2x→ which is coplanar with a→ and b→ and perpendicular to vector b→ and a→.x→=7S)(−9,−1,1)

I→S,II→P,III→R,IV→Q(a) Let image be (α,β,γ) α−32=β−51=γ−71=−2(18)6 α=−9,β−1,γ=1 (b) Let foot be (α,β,γ) α−72=β−144=γ−5−1=−(65−2)21 α−72=−3⇒α=1, β−14=−12⇒β=2, γ−5=3⇒γ=8 (c) R→×B→=C→×B→ R→−C→=λB→ R→=4i→−3j→+7k→+λ(i→+j→+k→) R→=(4+λ)i→+(λ−3)j→+(7+λ)k→ R→.A→=0⇒λ=−5 ∴R→=(−1,−8,2) (d) x→=λ(b→×(a→×b→))=λ((b→.b→)a→−(b→.a→)b→)=λ(5(−i→+j→+k→)+1(2i→+k→)) x→=λ(−3i→+5j→+6k→) Now a→.x→=7⇒3λ+5λ+6λ=7⇒λ=12 ∴2x→=−3i→+5j→+6k→

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Column – I Column – II
I) Image of the point $P (3, 5, 7)$ in the plane $2 x + y + z = 0$ P) $(1, 2, 8)$
II) Foot of the perpendicular from $(7, 14, 5)$ on to the plane $2 x + 4 y - z = 2$ Q) $(- 3, 5, 6)$
III) Let $\vec{A} = 2 \vec{i} + \vec{k}, \vec{B} = \vec{i} + \vec{j} + \vec{k}$ , $\vec{C} = 4 \vec{i} - 3 \vec{j} + 7 \vec{k}$ . If $\vec{R} \times \vec{B} = \vec{C} \times \vec{B}$ and $\vec{R} . \vec{A} = 0$ then coordinates of $\vec{R}$ are R) $(- 1, - 8, 2)$
IV) If $\vec{a} = - \vec{i} + \vec{j} + \vec{k}$ , $\vec{b} = 2 \vec{i} + \vec{k}$ then coordinates of a vector $2 \vec{x}$ which is coplanar with $\vec{a}$ and $\vec{b}$ and perpendicular to vector $\vec{b}$ and $\vec{a} . \vec{x} = 7$ S) $(- 9, - 1, 1)$

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$I \to P,II \to S,III \to R,IV \to Q$

$I \to S,II \to P,III \to R,IV \to Q$

$I \to S,II \to P,III \to Q,IV \to R$

$I \to P,II \to S,III \to Q,IV \to R$

answer is A.

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Detailed Solution

$I \to S,II \to P,III \to R,IV \to Q$

(a) Let image be $(α, β, γ)$
   $\frac{α - 3}{2} = \frac{β - 5}{1} = \frac{γ - 7}{1} = \frac{- 2 (18)}{6} α = - 9, β - 1, γ = 1$
(b) Let foot be $(α, β, γ)$
   $\frac{α - 7}{2} = \frac{β - 14}{4} = \frac{γ - 5}{- 1} = \frac{- (65 - 2)}{21} \frac{α - 7}{2} = - 3 \Rightarrow α = 1, β - 14 = - 12 \Rightarrow β = 2, γ - 5 = 3 \Rightarrow γ = 8$
(c) $\vec{R} \times \vec{B} = \vec{C} \times \vec{B}$
   $\vec{R} - \vec{C} = λ \vec{B} \vec{R} = 4 \vec{i} - 3 \vec{j} + 7 \vec{k} + λ (\vec{i} + \vec{j} + \vec{k}) \vec{R} = (4 + λ) \vec{i} + (λ - 3) \vec{j} + (7 + λ) \vec{k} \vec{R} . \vec{A} = 0 \Rightarrow λ = - 5 ∴ \vec{R} = (- 1, - 8, 2)$
(d) $\vec{x} = λ (\vec{b} \times (\vec{a} \times \vec{b})) = λ ((\vec{b} . \vec{b}) \vec{a} - (\vec{b} . \vec{a}) \vec{b}) = λ (5 (- \vec{i} + \vec{j} + \vec{k}) + 1 (2 \vec{i} + \vec{k}))$
   $\vec{x} = λ (- 3 \vec{i} + 5 \vec{j} + 6 \vec{k})$
Now $\vec{a} . \vec{x} = 7 \Rightarrow 3 λ + 5 λ + 6 λ = 7 \Rightarrow λ = \frac{1}{2}$
$∴ 2 \vec{x} = - 3 \vec{i} + 5 \vec{j} + 6 \vec{k}$

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