Respuesta :
Answer:
To convert the vector equation to cartesian form using the vector product, we can take the cross product of the two vectors inside the parentheses and use it as a normal vector to a plane, which contains the given point and is parallel to the direction of the third vector. Then, we can find the intersection point of this plane with the line passing through the given point and parallel to the third vector.
The cross product of (3i + j - 2k) and (-i + 2j + k) can be found as follows:
(3i + j - 2k) × (-i + 2j + k)
= (2j + 7k)i - 5j - 11k
Therefore, the equation of the plane containing the point (1, -5, 4) and parallel to the vector (2j + 7k)i - 5j - 11k is:
(2j + 7k)(x - 1) - 5(y + 5) - 11(z - 4) = 0
Expanding this equation gives:
2x + 7z - 15y - 51 = 0
Now, we need to find the intersection point of this plane with the line passing through the point (1, -5, 4) and parallel to the vector (-i + 2j + k). The parametric equations of this line are:
x = 1 - t
y = -5 + 2t
z = 4 + t
Substituting these into the equation of the plane, we get:
2(1 - t) + 7(4 + t) - 15(-5 + 2t) - 51 = 0
Simplifying this equation yields:
23t = 110
Therefore, t = 110/23, and the intersection point is:
x = 1 - 110/23 = -87/23
y = -5 + 2(110/23) = 75/23
z = 4 + 110/23 = 197/23
Hence, the cartesian form of the vector equation is:
(x, y, z) = (-87/23, 75/23, 197/23) + λ(3, 1, -2) + µ(-1, 2, 1)
Step-by-step explanation:
Answer:
[tex]5x-y+7z-38=0[/tex]
or
[tex]5x-y+7z=38[/tex]
Step-by-step explanation:
Given vector equation:
[tex]\textbf{r} = \textbf{i} - 5\textbf{j} + 4\textbf{k} + \lambda(3\textbf{i} + \textbf{j} - 2\textbf{k}) + \mu(-\textbf{i} + 2\textbf{j} + \textbf{k})[/tex]
The vector 3i + j - 2k is perpendicular to n.
The vector -i + 2j + k is perpendicular to n.
[tex]\textsf{So}\;\;\textbf{n}=\left|\begin{array}{ccc}\textbf{i}&\textbf{j}&\textbf{k}\\3&1&-2\\-1&2&1\end{array}\right|[/tex]
[tex]=\textbf{i}\left|\begin{array}{rr}1&-2\\2&1\end{array}\right|-\textbf{j}\left|\begin{array}{rr}3&-2\\-1&1\end{array}\right|+\textbf{k}\left|\begin{array}{rr}3&1\\-1&2\end{array}\right|[/tex]
[tex]=\textbf{i}(1 \cdot 1 - (-2) \cdot 2)-\textbf{j}(3 \cdot 1-(-2) \cdot (-1))+\textbf{k}(3 \cdot 2-1 \cdot (-1))[/tex]
[tex]=\textbf{i}(1 +4)-\textbf{j}(3-2)+\textbf{k}(6+1)[/tex]
[tex]=5\textbf{i}-\textbf{j}+7\textbf{k}[/tex]
So the equation of the plane written in Cartesian form is:
[tex]5x-y+7z+d=0[/tex]
Substituting (1, -5, 4) gives:
[tex]5(1)-(-5)+7(4)+d=0[/tex]
[tex]5+5+28+d=0[/tex]
[tex]38+d=0[/tex]
[tex]d=-38[/tex]
Therefore, the given vector equation in Cartesian form is:
[tex]5x-y+7z-38=0[/tex]
or
[tex]5x-y+7z=38[/tex]
