Let alpha x+beta y+gamma z=1 be the equation | vedclass

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(B) The normal vector $\vec{n}$ of the plane is the direction vector of the line joining $(1, 2, 3)$ and $(-2, 3, 5)$.$\vec{n} = (-2-1)\hat{i} + (3-2)

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The normal vector $\vec{n}$ of the plane is the direction vector of the line joining $(1, 2, 3)$ and $(-2, 3, 5)$.$\vec{n} = (-2-1)\hat{i} + (3-2)\hat{j} + (5-3)\hat{k} = -3\hat{i} + \hat{j} + 2\hat{k}$.Since the plane is perpendicular to this line,the normal vector is $\vec{n} = -3\hat{i} + \hat{j} + 2\hat{k}$.The equation of the plane passing through $(x_0, y_0, z_0)$ with normal $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.Substituting the point $(3, -2, 5)$ and normal vector $(-3, 1, 2)$:$-3(x-3) + 1(y+2) + 2(z-5) = 0$.$-3x + 9 + y + 2 + 2z - 10 = 0$.$-3x + y + 2z = -1$.Multiplying by $-1$ to match the form $\alpha x + \beta y + \gamma z = 1$:$3x - y - 2z = 1$.Comparing with $\alpha x + \beta y + \gamma z = 1$,we get $\alpha = 3$,$\beta = -1$,and $\gamma = -2$.The product $\alpha \beta \gamma = (3)(-1)(-2) = 6$.

Let $\alpha x+\beta y+\gamma z=1$ be the equation of a plane passing through the point $(3, -2, 5)$ and perpendicular to the line joining the points $(1, 2, 3)$ and $(-2, 3, 5)$. Then the value of $\alpha \beta \gamma$ is equal to $..........$.

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