If the equation of a plane passing through A( | vedclass

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Question

(A) Step $1$: Since the plane is parallel to the $z$-axis, its normal vector $\vec{n}$ is perpendicular to the $z$-axis vector $\vec{k} = (0, 0, 1)$.

Vedclass · Accepted Answer

$p = -1, q = 5$

Vedclass · Answer

(A) Step $1$: Since the plane is parallel to the $z$-axis, its normal vector $\vec{n}$ is perpendicular to the $z$-axis vector $\vec{k} = (0, 0, 1)$. Thus, the $z$-component of the normal vector is $0$. The equation of the plane is $3x - 2y - q = 0$, which matches this condition.Step $2$: The plane passes through $A(1, p, 2)$. Substituting these coordinates into the equation: $3(1) - 2(p) - q = 0 \implies 3 - 2p - q = 0$.Step $3$: The plane passes through $B(3, 2, 4)$. Substituting these coordinates into the equation: $3(3) - 2(2) - q = 0 \implies 9 - 4 - q = 0 \implies 5 - q = 0 \implies q = 5$.Step $4$: Substitute $q = 5$ into the equation from Step $2$: $3 - 2p - 5 = 0 \implies -2p - 2 = 0 \implies p = -1$.Step $5$: Thus, $p = -1$ and $q = 5$.

If the equation of a plane passing through $A(1, p, 2)$ and $B(3, 2, 4)$ and parallel to the $z$-axis is $3x - 2y - q = 0$, then:

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