17.  Find the equation of the plane which contains the line of intersection of the planes \overrightarrow{r}.(\widehat{i}+2\widehat{j}+3\widehat{k})-4=0,\overrightarrow{r}.\left ( 2\widehat{i}+\widehat{j}-\widehat{k} \right )+5=0and which is perpendicular to the plane \overrightarrow{r}.(5\widehat{i}+3\widehat{j}-6\widehat{k})+8=0

Answers (1)
D Divya Prakash Singh

The equation of the plane passing through the line of intersection of the given plane in  \overrightarrow{r}.(\widehat{i}+2\widehat{j}+3\widehat{k})-4=0,\overrightarrow{r}.\left ( 2\widehat{i}+\widehat{j}-\widehat{k} \right )+5=0

\left [ \vec{r}.(\widehat{i}+2\widehat{j}+3\widehat{k})-4 \right ]+\lambda\left [ \vec{r}.\left ( 2\widehat{i}+\widehat{j}-\widehat{k} \right )+5 \right ] = 0

\vec{r}.[(2\lambda+1)\widehat{i}+(\lambda+2)\widehat{j}+(3-\lambda)\widehat{k}]+(5\lambda-4)= 0       ,,,,,,,,,,,,,(1)

The plane in equation (1) is perpendicular to the plane, 

\vec{r}.\left ( 5\widehat{i}+3\widehat{j}-6\wideahat{k} \right ) +8 = 0

Therefore 5(2\lambda+1)+3(\lambda+2) -6(3-\lambda) = 0

\implies 19\lambda -7 = 0

\implies \lambda = \frac{7}{19}

Substituting \lambda = \frac{7}{19} in equation (1), we obtain

\implies \vec{r}.\left [ \frac{33}{19}\widehat{i}+\frac{45}{19}\widehat{j}+\frac{50}{19}\widehat{k} \right ] -\frac{41}{19} = 0

\implies \vec{r}.(33\widehat{i}+45\widehat{j}+50\widehat{k}) -41 = 0                     .......................(4)

So, this is the vector equation of the required plane.

The Cartesian equation of this plane can be obtained by substituting \implies \vec{r}= (x\widehat{i}+y\widehat{j}+z\widehat{k}) in equation (1).

(x\widehat{i}+y\widehat{j}+z\widehat{k}).(33\widehat{i}+45\widehat{j}+50\widehat{k}) -41 = 0

Therefore we get the answer 33x+45y+50z -41 = 0

 

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