Question

Let ${\displaystyle \overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k},\overrightarrow{b}=\hat{i}+\hat{j}+\hat{k}}$ and ${\displaystyle \overrightarrow{c}}$ be a vector such that ${\displaystyle \overrightarrow{a}+(\overrightarrow{b}\times \overrightarrow{c})=\overrightarrow{0}}$ and ${\displaystyle \overrightarrow{b}.\overrightarrow{c}}$ = 5. Then, the value of ${\displaystyle 3(\overrightarrow{c}.\overrightarrow{a})}$ is equal to ______.

विकल्प

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Answer 1

Let ${\displaystyle \overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k},\overrightarrow{b}=\hat{i}+\hat{j}+\hat{k}}$ and ${\displaystyle \overrightarrow{c}}$ be a vector such that ${\displaystyle \overrightarrow{a}+(\overrightarrow{b}\times \overrightarrow{c})=\overrightarrow{0}}$ and ${\displaystyle \overrightarrow{b}.\overrightarrow{c}}$ = 5. Then, the value of ${\displaystyle 3(\overrightarrow{c}.\overrightarrow{a})}$ is equal to 0.

Explanation:

${\displaystyle \overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k}}$

${\displaystyle \overrightarrow{b}=\hat{i}+\hat{j}+\hat{k}}$

Now, ${\displaystyle \overrightarrow{a}.\overrightarrow{b}}$ = 1 – 2 + 3 = 2 ...(i)

Also, given ${\displaystyle \overrightarrow{a}+(\overrightarrow{b}\times \overrightarrow{c})}$ = 0

⇒ ${\displaystyle \overrightarrow{a}=-(\overrightarrow{b}\times \overrightarrow{c})}$

⇒ ${\displaystyle \overrightarrow{a}.\overrightarrow{b}=-(\overrightarrow{b}\times \overrightarrow{c}).\overrightarrow{b}}$ ...(ii)

Equation (i) and equation (ii) are contradicting.