A plane pi passing through the points 2 hat{i | vedclass

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Question

(A) Let the points be $P_1(2, -3, 0)$ and $P_2(3, 0, 4)$. The vector $\vec{v_1} = P_2 - P_1 = (3-2)\hat{i} + (0-(-3))\hat{j} + (4-0)\hat{k} = \hat{i}

Vedclass · Accepted Answer

$31$

Vedclass · Answer

(A) Let the points be $P_1(2, -3, 0)$ and $P_2(3, 0, 4)$. The vector $\vec{v_1} = P_2 - P_1 = (3-2)\hat{i} + (0-(-3))\hat{j} + (4-0)\hat{k} = \hat{i} + 3\hat{j} + 4\hat{k}$.Given the plane is parallel to $\vec{v_2} = 2\hat{i} + 3\hat{j} - 4\hat{k}$.The normal to the plane $\vec{n} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 3 & 4 \ 2 & 3 & -4 \end{vmatrix} = \hat{i}(-12-12) - \hat{j}(-4-8) + \hat{k}(3-6) = -24\hat{i} + 12\hat{j} - 3\hat{k}$.Dividing by $-3$, we get the normal vector $\vec{n} = 8\hat{i} - 4\hat{j} + \hat{k}$.The equation of the plane is $8(x-2) - 4(y+3) + 1(z-0) = 0 \Rightarrow 8x - 4y + z = 28$.The line joining $A(1, 2, 0)$ and $B(0, 1, -2)$ has direction vector $\vec{d} = B - A = -\hat{i} - \hat{j} - 2\hat{k}$.The line equation is $\frac{x-1}{-1} = \frac{y-2}{-1} = \frac{z-0}{-2} = k \Rightarrow x = 1-k, y = 2-k, z = -2k$.Substituting into the plane equation: $8(1-k) - 4(2-k) + (-2k) = 28 \Rightarrow 8 - 8k - 8 + 4k - 2k = 28 \Rightarrow -6k = 28 \Rightarrow k = -\frac{14}{3}$.Then $a = 1 - (-14/3) = 17/3$, $b = 2 - (-14/3) = 20/3$, $c = -2(-14/3) = 28/3$.Thus, $a+b+2c = \frac{17+20+56}{3} = \frac{93}{3} = 31$.

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