Question

find an equation for the plane that passes through the point (1, -9, 8) and is orthogonal to the line \\( \mathbf{r}(t) = \langle 8, 1, -9 \
angle + t\langle -9, -8, -3 \
angle \\). give your answer in the form \\( ax + by + cz = d \\) (with \\( a = 9 \\)). \\( a = \\) blank, \\( b = \\) blank, \\( c = \\) blank, \\( d = \\) blank. submit answer, next item.

Explanation:

Step1: Identify the normal vector

The line has direction vector $\mathbf{v}=\langle -9, -8, -3
angle$. Since the plane is orthogonal to the line, the direction vector of the line is the normal vector $\mathbf{n}$ of the plane. Given $a = 9$, we need to scale the normal vector to make $a = 9$. The original normal vector component for $x$ is $-9$, so we multiply the normal vector by $-1$ to get $\mathbf{n}=\langle 9, 8, 3
angle$. So $a = 9$, $b = 8$, $c = 3$.

Step2: Find $d$ using the point

The plane passes through the point $(x_0, y_0, z_0)=(1, -9, 8)$. The equation of the plane is $ax + by + cz = d$. Substitute $a = 9$, $b = 8$, $c = 3$, $x = 1$, $y = -9$, $z = 8$ into the equation:

$$ d=9(1)+8(-9)+3(8) $$

Calculate each term: $9(1)=9$, $8(-9)= -72$, $3(8)=24$. Then sum them: $9 - 72 + 24=9 + 24 - 72=33 - 72=-39$.

Answer:

$a = 9$, $b = 8$, $c = 3$, $d = -39$