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CBSE Class 12 Math 2022 (Term II) Outside Delhi Set 2 Solved Paper

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SECTION - B
Question Nos. 7 to 10 carry 3 marks each.
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Question : 2 of 4
Marks: +1, -0
If a→,b→,c→\overset{\rightarrow}{a}, \overset{\rightarrow}{b}, \overset{\rightarrow}{c} are three vectors such that a→⋅b→=a→⋅c→\overset{\rightarrow}{a}\cdot \overset{\rightarrow}{b}=\overset{\rightarrow}{a}\cdot \overset{\rightarrow}{c} and a→×b→=\overset{\rightarrow}{a}\times \overset{\rightarrow}{b}= a→×c→,a→≠0\overset{\rightarrow}{a}\times \overset{\rightarrow}{c}, \overset{\rightarrow}{a}\neq 0, then show that b→=c→\overset{\rightarrow}{b}=\overset{\rightarrow}{c}.
OR
If ∣a⃗∣=3,∣b⃗∣=5,∣c⃗∣=4|\vec{a}|=3,|\vec{b}|=5,|\vec{c}|=4 and a⃗+b⃗+c⃗=0→\vec{a}+\vec{b}+\vec{c}=\overset{\rightarrow}{0}, then find the value of (a→⋅b→+b→⋅c→+c→⋅a→)\left(\overset{\rightarrow}{a}\cdot \overset{\rightarrow}{b}+\overset{\rightarrow}{b}\cdot \overset{\rightarrow}{c}+\overset{\rightarrow}{c}\cdot \overset{\rightarrow}{a}\right)
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