The vector equation of the plane passing through
Question:

The vector equation of the plane passing through

the intersection of the planes $\overrightarrow{\mathrm{r}} .(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=1$ and

$\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}-2 \hat{\mathrm{j}})=-2$, and the point $(1,0,2)$ is :

 

  1. $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=\frac{7}{3}$

  2. $\overrightarrow{\mathrm{r}} \cdot(3 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=7$

  3. $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=7$

  4. $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=\frac{7}{3}$


Correct Option: , 3

Solution:

$\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=1$

$\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}-2 \hat{\mathrm{j}})=-2$

point $(1,0,2)$

$\mathrm{Eq}^{\mathrm{n}}$ of plane

$\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})-1+\lambda\{\mathrm{r} \cdot(\hat{\mathrm{i}}-2 \hat{\mathrm{j}})+2\}=0$

$\overrightarrow{\mathrm{r}} \cdot\{\hat{\mathrm{i}}(1+\lambda)+\hat{\mathrm{j}}(1-2 \lambda)+\hat{\mathrm{k}}(1)\}-1+2 \lambda=0$

Point $\hat{\mathrm{i}}+0 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}=\overrightarrow{\mathrm{r}}$

$\therefore(\hat{\mathrm{i}}+2 \hat{\mathrm{k}})\{\hat{\mathrm{i}}(1+\lambda)+\hat{\mathrm{j}}(1-2 \lambda)+\hat{\mathrm{k}}(1)\}-1+2 \lambda=0$

$1+\lambda+2-1+2 \lambda=0$

$\lambda=-\frac{2}{3}$

$\therefore \quad \overrightarrow{\mathrm{r}} \cdot\left[\hat{\mathrm{i}}\left(\frac{1}{3}\right)+\hat{\mathrm{j}}\left(\frac{7}{3}\right)+\hat{\mathrm{k}}\right]=\frac{7}{3}$

$\mathrm{r} \cdot[\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}]=7$

Ans. 3

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