Question

the cross product of two 3d vectors will always produce a vector that is opposite orthogonal parallel a scalar multiple equal in magnitude to the two original vectors. classify the conic section by the general form of the equation given. enter one of the following in the box below: circle, ellipse, parabola, hyperbola (9x^2 + 162x + 4y^2 + 8y + 732 = 0) answer:

Explanation:

First Question (Vector Cross Product)

By the definition of the cross product in 3D vector mathematics, the cross product of two vectors $\vec{u}$ and $\vec{v}$ (denoted as $\vec{u} \times \vec{v}$) results in a vector that is perpendicular (orthogonal) to both $\vec{u}$ and $\vec{v}$. The other options do not fit: "opposite" implies direction reversal (not a property of cross - product result's relation to originals), "parallel" would mean it's in the same or opposite direction of one of the vectors (which is not true), "a scalar multiple" is a property of parallel vectors (not cross - product), and "equal in magnitude" is not a defining relation of the cross - product vector to the original vectors.

Second Question (Conic Section Classification)

Step1: Complete the square for x and y terms

Given the equation $9x^{2}+162x + 4y^{2}+8y+732 = 0$.
For the x - terms:
Factor out the coefficient of $x^{2}$: $9(x^{2}+18x)+4y^{2}+8y+732 = 0$.
Complete the square inside the parentheses for $x$: $x^{2}+18x=(x + 9)^{2}-81$. So, $9((x + 9)^{2}-81)+4y^{2}+8y+732 = 0$.
For the y - terms:
Factor out the coefficient of $y^{2}$: $9((x + 9)^{2}-81)+4(y^{2}+2y)+732 = 0$.
Complete the square inside the parentheses for $y$: $y^{2}+2y=(y + 1)^{2}-1$. So, $9((x + 9)^{2}-81)+4((y + 1)^{2}-1)+732 = 0$.

Step2: Simplify the equation

Expand the equation:
$9(x + 9)^{2}-729+4(y + 1)^{2}-4 + 732 = 0$.
Combine like terms:
$9(x + 9)^{2}+4(y + 1)^{2}-729-4 + 732 = 0$.
$9(x + 9)^{2}+4(y + 1)^{2}-1 = 0$.
Rearrange to the standard form:
$9(x + 9)^{2}+4(y + 1)^{2}=1$.
Divide both sides by 1: $\frac{(x + 9)^{2}}{\frac{1}{9}}+\frac{(y + 1)^{2}}{\frac{1}{4}}=1$.

Since the equation is in the form $\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}} = 1$ (where $h=-9,k = - 1,a^{2}=\frac{1}{9},b^{2}=\frac{1}{4}$) with $a
eq b$ and both $x^{2}$ and $y^{2}$ terms having positive coefficients, the conic section is an ellipse.

Answer:

orthogonal