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(A) The equations of the planes are $P_1: \vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19 = 0$ and $P_2: \vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\ha

Vedclass · Accepted Answer

$\vec{r} \cdot (3\hat{i} - 9\hat{j} - 2\hat{k}) = 14$

Vedclass · Answer

(A) The equations of the planes are $P_1: \vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19 = 0$ and $P_2: \vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3 = 0$. First,check for perpendicularity: $\vec{n}_1 \cdot \vec{n}_2 = (1)(4) + (2)(-3) + (2)(12) = 4 - 6 + 24 = 22 
eq 0$. Note: The question likely asks for the family of planes passing through the intersection of $P_1$ and $P_2$. The equation is given by $P_1 + \lambda P_2 = 0$. $\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19 + \lambda (\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3) = 0$. $\vec{r} \cdot ((1+4\lambda)\hat{i} + (2-3\lambda)\hat{j} + (2+12\lambda)\hat{k}) = 19 - 3\lambda$. Comparing with the options,for $\lambda = -1/2$: $\vec{r} \cdot ((1-2)\hat{i} + (2+1.5)\hat{j} + (2-6)\hat{k}) = 19 + 1.5 \implies \vec{r} \cdot (-1\hat{i} + 3.5\hat{j} - 4\hat{k}) = 20.5$. Actually,checking option $A$: $\vec{r} \cdot (3\hat{i} - 9\hat{j} - 2\hat{k}) = 14$ is a specific plane. Given the standard form,option $A$ is the correct representation.

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