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(A) Let the points be $A(-1, -2, 0)$ and $B(2, 3, 5)$. The vector $\vec{AB} = (2 - (-1))i + (3 - (-2))j + (5 - 0)k = 3i + 5j + 5k$.The plane is parall

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$r \cdot (-30i + 13j + 5k) = 4$

Vedclass · Answer

(A) Let the points be $A(-1, -2, 0)$ and $B(2, 3, 5)$. The vector $\vec{AB} = (2 - (-1))i + (3 - (-2))j + (5 - 0)k = 3i + 5j + 5k$.The plane is parallel to the line with direction vector $\vec{v} = 2i + 5j - k$.The normal vector $\vec{n}$ to the plane is given by the cross product of $\vec{AB}$ and $\vec{v}$:$\vec{n} = \vec{AB} 	imes \vec{v} = \begin{vmatrix} i & j & k \ 3 & 5 & 5 \ 2 & 5 & -1 \end{vmatrix}$$= i(-5 - 25) - j(-3 - 10) + k(15 - 10) = -30i + 13j + 5k$.The equation of the plane passing through $(-1, -2, 0)$ is:$-30(x + 1) + 13(y + 2) + 5(z - 0) = 0$$-30x - 30 + 13y + 26 + 5z = 0$$-30x + 13y + 5z = 4$.In vector form,this is $r \cdot (-30i + 13j + 5k) = 4$.

The equation of the plane passing through the points $(-1, -2, 0)$ and $(2, 3, 5)$ and parallel to the line $r = -3j + k + \lambda(2i + 5j - k)$ is

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