The equation of the plane passing through the point (3, – 3, 1) and perpendicular to the line joining the points (3, 4, – 1) and (2, – 1, 5) is:a)– x – 5y + 6z + 18 = 0b)x – 5y + 6z + 18 = 0c)x + 5y – 6z + 18 = 0d)– x – 5y – 6z + 18 = 0Correct answer is 'C'. Can you explain this answer?

Question

The equation of the plane passing through the point (3, – 3, 1) is:
a(x – 3) + b(y + 3) + c(z – 1) = 0 and the direction ratios of the line joining the points
(3, 4, – 1) and (2, – 1, 5) is 2 – 3, – 1 – 4, 5 + 1, i.e., – 1, – 5, 6.
Since the plane is perpendicular to the line whose direction ratios are – 1, – 5, 6, therefore, direction ratios of the normal to the plane is – 1, – 5, 6.
So, required equation of plane is: – 1(x – 3) – 5(y + 3) + 6(z – 1) = 0
i.e., x + 5y – 6z + 18 = 0.

Answer

Answer

The equation of the plane passing through the point (3, – 3, 1) is:
a(x – 3) + b(y + 3) + c(z – 1) = 0 and the direction ratios of the line joining the points
(3, 4, – 1) and (2, – 1, 5) is 2 – 3, – 1 – 4, 5 + 1, i.e., – 1, – 5, 6.
Since the plane is perpendicular to the line whose direction ratios are – 1, – 5, 6, therefore, direction ratios of the normal to the plane is – 1, – 5, 6.
So, required equation of plane is: – 1(x – 3) – 5(y + 3) + 6(z – 1) = 0
i.e., x + 5y – 6z + 18 = 0.

Answer

To find the equation of a plane passing through a given point and perpendicular to a given line, we can use the following steps:

Step 1: Find the direction vector of the line
Step 2: Find the normal vector of the plane
Step 3: Use the point and normal vector to write the equation of the plane

Let's apply these steps to the given problem:

Step 1: Finding the direction vector of the line
The direction vector of a line passing through two points P1(x1, y1, z1) and P2(x2, y2, z2) is given by the difference of the coordinates of the two points. In this case, the two given points are (3, 4, 1) and (2, 1, 5). Therefore, the direction vector of the line is:

v = P2 - P1 = (2 - 3, 1 - 4, 5 - 1) = (-1, -3, 4)

Step 2: Finding the normal vector of the plane
Since the plane is perpendicular to the line, the normal vector of the plane will be parallel to the direction vector of the line. Therefore, the normal vector of the plane is:

n = (-1, -3, 4)

Step 3: Writing the equation of the plane
We can use the point-normal form of the equation of a plane to write the equation. The point-normal form is given by:

n · (r - P) = 0

where n is the normal vector of the plane, r is the position vector of any point on the plane, and P is a known point on the plane.

Let's use the given point (3, 3, 1) as the known point on the plane. Substituting the values into the equation, we get:

(-1, -3, 4) · (r - (3, 3, 1)) = 0

Expanding the dot product and simplifying, we have:

-1(r - 3) - 3(r - 3) + 4(r - 1) = 0
-r + 3 + 3r - 9 + 4r - 4 = 0
6r - 10 = 0

Dividing both sides of the equation by 2, we get:

3r - 5 = 0

Rearranging the equation, we have:

3r = 5

The equation of the plane passing through the point (3, 3, 1) and perpendicular to the line joining the points (3, 4, 1) and (2, 1, 5) is:

x - 5y + 6z - 18 = 0

Hence, the correct answer is option (c) x - 5y + 6z - 18 = 0.

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