If a → = 2 i ^ + k ^ ,   b → =   i ^ + j ^ + k ^ ,   c → = 4 i ^ - 3 j ^ +…

If a→=2i^+k^, b→= i^+j^+k^, c→=4i^-3j^+7k^ , then the vector r→ satisfying r→×b→= c→×b→ and r→ .a→=0 is
  1. i^+8j^+2k^
  2. i^-8j^+2k^
  3. i^-8j^-2k^
  4. -i^-8j^+2k^

Solution

Given \(\vec{a}=2 \hat{i}+\hat{k}, \quad \vec{b}=\hat{i}+\hat{j}+\hat{k}, \quad \vec{c}=4 \hat{i}-3 \hat{j}+7 \hat{k}\) Conditions: 1. \(\vec{F} \times \vec{b}=\vec{c} \times \vec{b}\) 2. \(\vec{F} \cdot \vec{d}=0\) \(\vec{r} \times \vec{b}=\vec{c} \times \vec{b} \Rightarrow(\vec{r}-\vec{c}) \times \vec{b}=0\) So, \(\vec{r}-\vec{c} \mid \vec{b}\) Hence \(\vec{r}=\vec{c}+\vec{\lambda} \vec{b}\) \(\vec{F}=(4,-3,7)+\lambda(1,1,1)=(4+\lambda,-3+\lambda, 7+\lambda)\) \(\begin{gathered} \vec{f} \cdot \vec{a}=0 \\ \left(4+\lambda_1-3+\lambda, 7+\lambda\right) \cdot(2,0,1)=0 \\ 2(4+\lambda)+(7+\lambda)=0 \\ 8+2 \lambda+7+\lambda=0 \\ 15+3 \lambda=0 \Rightarrow \lambda=-5 \end{gathered}\) \(\vec{r}=(4-5,-3-5,7-5)=(-1,-8,2)\) Final Answer \(r=-i-8 j+2 k\)

Asked in: MHT CET Full Test 7

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