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Mathematics Question on Vector Algebra

If a→\overrightarrow{a} and b→ \overrightarrow{b} are vectors such that ∣a→+b→∣=29| \overrightarrow{a}+\overrightarrow{b} |=\sqrt{29} and a→×(2i^+3j^+4k^)=(2i^+3j^+4k^)×b→,\overrightarrow{a}\times(2\widehat{i}+3\widehat{j}+4\widehat{k})=(2\widehat{i}+3\widehat{j}+4\widehat{k})\times\overrightarrow{b}, then a possible value of (a→+b→)(−7i^+2j^+3k^is(\overrightarrow{a}+\overrightarrow{b})(-7\widehat{i}+2\widehat{j}+3\widehat{k} is

A

0

B

3

C

4

D

8

Answer

4

Explanation

Solution

Plan If a→×b→=a→×c→\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{a}\times\overrightarrow{c}
⇒a→×b→−a→×c→=0⇒a→×(b→−c→)\Rightarrow \overrightarrow{a}\times\overrightarrow{b}-\overrightarrow{a}\times\overrightarrow{c}=0 \Rightarrow \overrightarrow{a}\times(\overrightarrow{b}-\overrightarrow{c})=0
i.e.a→∣∣(b→−c→)orb→−c→=λa→i.e. \overrightarrow{a} || (\overrightarrow{b}-\overrightarrow{c})or \overrightarrow{b}-\overrightarrow{c}=\lambda \overrightarrow{a}
Here,a→×(2i^+3j^+4k^)=(2i^+3j^+4k^)×b→Here, \overrightarrow{a}\times(2\widehat{i}+3\widehat{j}+4\widehat{k})=(2\widehat{i}+3\widehat{j}+4\widehat{k})\times\overrightarrow{b}
⇒a→×(2i^+3j^+4k^)−(2i^+3j^+4k^)×b→=0\Rightarrow \overrightarrow{a} \times(2\widehat{i}+3\widehat{j}+4\widehat{k})-(2\widehat{i}+3\widehat{j}+4\widehat{k})\times\overrightarrow{b}=0
⇒(a→+b→)×(2i^+3j^+4k^)=0\Rightarrow (\overrightarrow{a}+\overrightarrow{b})\times(2\widehat{i}+3\widehat{j}+4\widehat{k})=0
\Rightarrow \overrightarrow{a}+\overrightarrow{b}=\lambda (2\widehat{i}+3\widehat{j}+4\widehat{k})\hspace25mm ...(i)
Since, ∣a→+b→∣=29⇒±λ4+9+16=29| \overrightarrow{a} +\overrightarrow{b} |=\sqrt{29} \Rightarrow \pm \lambda\sqrt{4+9+16}=\sqrt{29}
⇒λ=±1\Rightarrow \lambda=\pm1
\therefore \hspace25mm \overrightarrow{a}+\overrightarrow{b}=\pm(2\widehat{i}+3\widehat{j}+4\widehat{k})
Now, (a→+b→).(−7i^+2j^+3k^)=±(−14+6+12)=±4(\overrightarrow{a}+\overrightarrow{b}).(-7\widehat{i}+2\widehat{j}+3\widehat{k})=\pm(-14+6+12)=\pm4