Let a→=2i^+3j^+4k^, b→=i^-2j^+k^ and c→=i^+j^-k^. If r→×a→=b→ and r→⋅c→=3, then the value of r→ is equal to

Correct answer: (A) √155. Given, r→×a→=b→ Taking right cross product with c→, we get, r→×a→×c→=b→×c→⇒r→⋅c→a→-a→⋅c→r→=b→×c→ ⇒3a→-r→=b→×c→⇒r→=32i^+3j^+4k^-i^+2j^+3k^ ⇒r→=5i^+7j^+9k^⇒r→=√155

Q17 / 100 · NTA Abhyas (JEE Main) 2020 · Mathematics · Vector Algebra

Let $\vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}, \vec{b} = \hat{i} - 2 \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} + \hat{j} - \hat{k}$ . If $\vec{r} \times \vec{a} = \vec{b}$ and $\vec{r} \cdot \vec{c} = 3$ , then the value of $|\vec{r}|$ is equal to

Correct answer
A
$\sqrt{155}$
B
$\sqrt{17}$
C
$2 \sqrt{17}$
D
$3$

Solution

Given,

\vec{r} \times \vec{a} = \vec{b}

Taking right cross product with

\vec{c}

, we get,

(\vec{r} \times \vec{a}) \times \vec{c} = \vec{b} \times \vec{c} \Rightarrow (\vec{r} \cdot \vec{c}) \vec{a} - (\vec{a} \cdot \vec{c}) \vec{r} = \vec{b} \times \vec{c}

\Rightarrow 3 \vec{a} - \vec{r} = \vec{b} \times \vec{c} \Rightarrow \vec{r} = 3 (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) - (\hat{i} + 2 \hat{j} + 3 \hat{k})

\Rightarrow \vec{r} = 5 \hat{i} + 7 \hat{j} + 9 \hat{k} \Rightarrow |\vec{r}| = \sqrt{155}

Hence, the correct answer is option: 'A'.

Let a→=2i^+3j^+4k^, b→=i^-2j^+k^ and c→=i^+j^-k^. If r→×a→=b→ and r→⋅c→=3, then the value of r→ is equal to

Solution

More Vector Algebra Questions

Let $\vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}, \vec{b} = \hat{i} - 2 \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} + \hat{j} - \hat{k}$ . If $\vec{r} \times \vec{a} = \vec{b}$ and $\vec{r} \cdot \vec{c} = 3$ , then the value of $|\vec{r}|$ is equal to