<p>The point of intersection of the plane \(\vec{r} \cdot (3\hat{i} - 5\hat{j} + 2\hat{k}) = 6\) with the straight line passing through the origin and perpendicular to the plane \(2x - y - z = 4\), is \((x_0, y_0, z_0)\). The value of \((2x_0 - 3y_0 + z_0)\), is:</p>
Step-by-Step Solution
Key Concept: The line perpendicular to plane 2x - y - z = 4 has direction vector (2, -1, -1) (the normal to the plane). Parametrize this line through origin as r(t) = t(2, -1, -1) and substitute into the given plane equation to find t, then compute the required linear combination.
Step 1: Find the direction vector of the line. Since the line is perpendicular to plane 2x - y - z = 4, its direction vector is the normal to this plane: d = (2, -1, -1). Step 2: Write the parametric equation of the line passing through origin with direction (2, -1, -1):
x = 2t, y = -t, z = -t Step 3: Substitute into the plane equation r · (3 î - 5 ĵ + 2 k̂ ) = 6:
3(2t) - 5(-t) + 2(-t) = 6
6t + 5t - 2t = 6
9t = 6
t = 2/3 Step 4: Find the intersection point:
x_0 = 2(2/3) = 4/3
y_0 = -(2/3) = -2/3
z_0 = -(2/3) = -2/3 Step 5: Calculate 2x_0 - 3y_0 + z_0:
= 2(4/3) - 3(-2/3) + (-2/3)
= 8/3 + 6/3 - 2/3
= 12/3 = 4 ∴ Answer: 4
Correct Answer: C