Find the distance of the point P(– 2, – 4, 7) from the point of intersection Q of the line \overrightarrow{r}=(3\widehat{i}-2\widehat{j}+6\widehat{k})+\lambda (2\widehat{i}-\widehat{j}+2\widehat{k}) and the plane \overrightarrow{r}.\left (\widehat{i}-\widehat{j}+\widehat{k} \right )=6  Also write the vector equation of the line PQ.

 

Answers (1)

Given :

Line: \overrightarrow{r}=(3\widehat{i}-2\widehat{j}+6\widehat{k})+\lambda (2\widehat{i}-\widehat{j}+2\widehat{k})

\overrightarrow{\mathrm{r}}=(3+2 \lambda) \hat{\mathrm{i}}+(-2-\lambda) \hat{\mathrm{j}}+(6+2 \lambda) \hat{\mathrm{k}}\\

Plane:  \overrightarrow{r}.\left (\widehat{i}-\widehat{j}+\widehat{k} \right )=6

Intersection Point:

\\ &[(3+2 \lambda) \hat{i}+(-2-\lambda) \hat{j}+(6+2 \lambda) \hat{k}] \cdot(\hat{i}-\hat{j}+\hat{k})=6

\\ 3+2 \lambda+2+\lambda+6+2 \lambda=6 \\ 5 \lambda = -5 \Rightarrow \lambda=-1

Point Q (1,-1,4)

Point P(-2,-4,7)

&\mathrm{PQ}= \sqrt{(1+2)^2+(-1+4)^2+(4-7)^2} = \sqrt{9+9+9 } =3 \sqrt{3}

\text { Line } \mathrm{PQ}: \overrightarrow{\mathrm{r}}=-2 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}+\mu(3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})

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