Match the following.LIST-ILIST-IIA)If a→,b→ and c→ are three mutually perpendicular vectors where|a→|=|b→|=2,|c→|=1,then [a→×b→ b→×c→ c→×a→] isP)-12B)If a→ and b→ are two unit vectors inclined at π3, then 16[a→ b→+(a→×b→) b→] isQ)0C)If b→and c→ are orthogonal unit vectors and b→×c→=a→then [a→+b→+c→ a→+b→ b→+c→] isR)16D)If [x→ y→ a→]=[x→ y→ b→]=[a→ b→ c→]=0 ,each vector being a non-zero vector, then [x→ y→ c→] isS)1 T)4

A) If a→ , b→ and c→ are mutually perpendicular, then [a→×b→ b→×c→ c→×a→]=[a→ b→ c→]2=(|a→||b→||c→|)2=16 B) Given a→ and b→ are two unit vectors, i.e., |a→|=|b→|=1 and angle between them is π3 . sinθ=|a→×b→||a→||b→|⇒sinπ3=|a→×b→|;32=|a→×b→|Now [a→ b→+a→×b→b→]=[a→ b→ b→]+[a→ a→×b→ b→]=0+[a→ a→×b→ b→ ]=(a→×b→).(b→×a→)=−|a→×b→|2=−34 C) If b→ and c→ are orthogonal, b→ . c→=0Also, it is given that b→×c→=a→Now [a→+b→+c→a→+b→b→+c→]=[a→a→+b→b→+c→]+[b→+c→a→+b→b→+c→] =[a→b→c→]=a→.(b→×c→]=a→ . a→=|a→|2=1(because a→ is a unit vector) D) [x→y→a→]=0Therefore, x→,y→ and a→ are coplanar. [x→,y→ b→]=0Therefore, x→,y→ and b→ are coplanar.Also, [a→ b→ c→]=0Therefore, a→,b→ and c→ are coplanar.From (i), (ii) and (iii)x→,y→ and c→ are coplanar. Therefore, [x→,y→ c→]=0

Match the following.LIST-ILIST-IIA)If a→,b→ and c→ are three mutually perpendicular vectors where|a→|=|b→|=2,|c→|=1,then [a→×b→ b→×c→ c→×a→] isP)-12B)If a→ and b→ are two unit vectors inclined at π3, then 16[a→ b→+(a→×b→) b→] isQ)0C)If b→and c→ are orthogonal unit vectors and b→×c→=a→then [a→+b→+c→ a→+b→ b→+c→] isR)16D)If [x→ y→ a→]=[x→ y→ b→]=[a→ b→ c→]=0 ,each vector being a non-zero vector, then [x→ y→ c→] isS)1 T)4

A) If a→ , b→ and c→ are mutually perpendicular, then [a→×b→ b→×c→ c→×a→]=[a→ b→ c→]2=(|a→||b→||c→|)2=16 B) Given a→ and b→ are two unit vectors, i.e., |a→|=|b→|=1 and angle between them is π3 . sinθ=|a→×b→||a→||b→|⇒sinπ3=|a→×b→|;32=|a→×b→|Now [a→ b→+a→×b→b→]=[a→ b→ b→]+[a→ a→×b→ b→]=0+[a→ a→×b→ b→ ]=(a→×b→).(b→×a→)=−|a→×b→|2=−34 C) If b→ and c→ are orthogonal, b→ . c→=0Also, it is given that b→×c→=a→Now [a→+b→+c→a→+b→b→+c→]=[a→a→+b→b→+c→]+[b→+c→a→+b→b→+c→] =[a→b→c→]=a→.(b→×c→]=a→ . a→=|a→|2=1(because a→ is a unit vector) D) [x→y→a→]=0Therefore, x→,y→ and a→ are coplanar. [x→,y→ b→]=0Therefore, x→,y→ and b→ are coplanar.Also, [a→ b→ c→]=0Therefore, a→,b→ and c→ are coplanar.From (i), (ii) and (iii)x→,y→ and c→ are coplanar. Therefore, [x→,y→ c→]=0

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Match the following.
LIST-I
LIST-II
A)
If $\vec{a}, \vec{b}$ and $\vec{c}$ are three mutually perpendicular vectors where $| \vec{a} | = | \vec{b} | = 2, | \vec{c} | = 1$ ,then $[\vec{a} \times \vec{b} \vec{b} \times \vec{c} \vec{c} \times \vec{a}]$ is
P)
-12
B)
If $\vec{a}$ and $\vec{b}$ are two unit vectors inclined at $\frac{π}{3}$ , then $16 [\vec{a} \vec{b} + (\vec{a} \times \vec{b}) \vec{b}]$ is
Q)
0
C)
If $\vec{b}$ and $\vec{c}$ are orthogonal unit vectors and $\vec{b} \times \vec{c} = \vec{a}$ then $[\vec{a} + \vec{b} + \vec{c} \vec{a} + \vec{b} \vec{b} + \vec{c}]$ is
R)
16
D)
If $[\vec{x} \vec{y} \vec{a}] = [\vec{x} \vec{y} \vec{b}] = [\vec{a} \vec{b} \vec{c}] = 0$ ,each vector being a non-zero vector, then $[\vec{x} \vec{y} \vec{c}]$ is
S)
1

T)
4

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

$A \to R; B \to P; C \to P; D \to Q$

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

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

answer is D.

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Detailed Solution

A) If $\vec{a}, \vec{b}$ and $\vec{c}$ are mutually perpendicular, then $[\vec{a} \times \vec{b} \vec{b} \times \vec{c} \vec{c} \times \vec{a}] = {[\vec{a} \vec{b} \vec{c}]}^{2} = {(| \vec{a} | | \vec{b} | | \vec{c} |)}^{2} = 16$
B) Given $\vec{a}$ and $\vec{b}$ are two unit vectors, i.e., $| \vec{a} | = | \vec{b} | = 1$ and angle between them is $\frac{π}{3}$ .
$\sin θ = \frac{| \vec{a} \times \vec{b} |}{| \vec{a} | | \vec{b} |} \Rightarrow \sin \frac{π}{3} = | \vec{a} \times \vec{b} |; \frac{\sqrt{3}}{2} = | \vec{a} \times \vec{b} |$
Now $[\vec{a} \vec{b} + \vec{a} \times \vec{b} \vec{b}] = [\vec{a} \vec{b} \vec{b}] + [\vec{a} \vec{a} \times \vec{b} \vec{b}] = 0 + [\vec{a} \vec{a} \times \vec{b} \vec{b}]$
$= (\vec{a} \times \vec{b}) . (\vec{b} \times \vec{a}) = - {| \vec{a} \times \vec{b} |}^{2} = - \frac{3}{4}$
C) If $\vec{b}$ and $\vec{c}$ are orthogonal, $\vec{b} . \vec{c} = 0$
Also, it is given that $\vec{b} \times \vec{c} = \vec{a}$
Now $[\vec{a} + \vec{b} + \vec{c} \vec{a} + \vec{b} \vec{b} + \vec{c}] = [\vec{a} \vec{a} + \vec{b} \vec{b} + \vec{c}] + [\vec{b} + \vec{c} \vec{a} + \vec{b} \vec{b} + \vec{c}]$
   $= [\vec{a} \vec{b} \vec{c}] = \vec{a} . (\vec{b} \times \vec{c}] = \vec{a} . \vec{a} = {| \vec{a} |}^{2} = 1$
(because $\vec{a}$ is a unit vector)
D) $[\vec{x} \vec{y} \vec{a}] = 0$
Therefore, $\vec{x}, \vec{y}$ and $\vec{a}$ are coplanar.
   $[\vec{x}, \vec{y} \vec{b}] = 0$
Therefore, $\vec{x}, \vec{y}$ and $\vec{b}$ are coplanar.
Also,    $[\vec{a} \vec{b} \vec{c}] = 0$
Therefore, $\vec{a}, \vec{b}$ and $\vec{c}$ are coplanar.
From (i), (ii) and (iii)
$\vec{x}, \vec{y}$ and $\vec{c}$ are coplanar. Therefore, $[\vec{x}, \vec{y} \vec{c}] = 0$

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