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(A) The equations of the given planes are:$\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4=0$  $(1)$$\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5=0$  $(2)$

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$\vec{r} \cdot(33 \hat{i}+45 \hat{j}+50 \hat{k})-41=0$

Vedclass · Answer

(A) The equations of the given planes are:$\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4=0$  $(1)$$\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5=0$  $(2)$The equation of the plane passing through the line of intersection of planes $(1)$ and $(2)$ is given by:$[\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4] + \lambda[\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5] = 0$$\vec{r} \cdot[(1+2\lambda) \hat{i} + (2+\lambda) \hat{j} + (3-\lambda) \hat{k}] + (5\lambda-4) = 0$  $(3)$Since this plane is perpendicular to the plane $\vec{r} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})+8=0$,the dot product of their normal vectors is zero:$5(1+2\lambda) + 3(2+\lambda) - 6(3-\lambda) = 0$$5 + 10\lambda + 6 + 3\lambda - 18 + 6\lambda = 0$$19\lambda - 7 = 0 \Rightarrow \lambda = \frac{7}{19}$Substituting $\lambda = \frac{7}{19}$ into equation $(3)$:$\vec{r} \cdot[(1 + \frac{14}{19}) \hat{i} + (2 + \frac{7}{19}) \hat{j} + (3 - \frac{7}{19}) \hat{k}] + (5(\frac{7}{19}) - 4) = 0$$\vec{r} \cdot[\frac{33}{19} \hat{i} + \frac{45}{19} \hat{j} + \frac{50}{19} \hat{k}] + (\frac{35-76}{19}) = 0$$\vec{r} \cdot(33 \hat{i} + 45 \hat{j} + 50 \hat{k}) - 41 = 0$

Find the equation of the plane which contains the line of intersection of the planes $\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4=0$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5=0$ and which is perpendicular to the plane $\vec{r} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})+8=0$.

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