Question

find a vector v parallel to the line of intersection of the planes ( y - 2x + 4z = 4 ) and ( y - 4x - 4z = 7 ).

Explanation:

Step1: Find normal vectors of planes

The first plane \( y - 2x + 4z = 4 \) can be written as \( -2x + y + 4z = 4 \), so its normal vector \( \mathbf{n_1}=\langle -2, 1, 4
angle \). The second plane \( y - 4x - 4z = 7 \) can be written as \( -4x + y - 4z = 7 \), so its normal vector \( \mathbf{n_2}=\langle -4, 1, -4
angle \).

Step2: Compute cross product of normal vectors

The direction vector of the line of intersection (and thus a vector parallel to it) is the cross product of \( \mathbf{n_1} \) and \( \mathbf{n_2} \). The cross product formula is \( \mathbf{n_1} \times \mathbf{n_2}=

$$\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-2&1&4\\-4&1&-4\end{vmatrix}$$

\).

Calculating the determinant:

• For \( \mathbf{i} \): \( (1)(-4)-(4)(1)= -4 - 4=-8 \)
• For \( \mathbf{j} \): \( -[(-2)(-4)-(4)(-4)] = -[8 + 16]=-24 \) (note the negative sign for \( \mathbf{j} \) component)
• For \( \mathbf{k} \): \( (-2)(1)-(-4)(1)= -2 + 4 = 2 \)

So \( \mathbf{n_1} \times \mathbf{n_2}=\langle -8, -24, 2
angle \). We can simplify this by dividing by 2: \( \langle -4, -12, 1
angle \) (or any scalar multiple, like \( \langle 4, 12, -1
angle \) by multiplying by -1).

Answer:

A vector \( \mathbf{v} \) parallel to the line is \( \langle -4, -12, 1
angle \) (or equivalent scalar multiples like \( \langle 4, 12, -1
angle \), \( \langle -8, -24, 2
angle \) etc.).