Question

The equation of the plane containing the $\frac{x}{2} = \frac{y}{3} =\frac{z}{4} $ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2} $ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3} $ is :

• A. x + 2y - 2z = 0
• B. x - 2y + z = 0
• C. 5x + 2y - 4z = 0
• D. 3x + 2y - 3z = 0

Answer 1

The Correct Option is B

To solve the problem, we need to find the equation of a plane that contains the line given by \(\frac{x}{2} = \frac{y}{3} =\frac{z}{4}\) and is perpendicular to another plane that contains the lines \(\frac{x}{3} = \frac{y}{4} = \frac{z}{2}\) and \(\frac{x}{4} = \frac{y}{2} = \frac{z}{3}\).

1. First, let's find the direction ratios of the given lines:
   • For the line represented by \(\frac{x}{2} = \frac{y}{3} =\frac{z}{4}\), the direction ratios are 2, 3, 4.
   • For the line \(\frac{x}{3} = \frac{y}{4} = \frac{z}{2}\), the direction ratios are 3, 4, 2.
   • For the line \(\frac{x}{4} = \frac{y}{2} = \frac{z}{3}\), the direction ratios are 4, 2, 3.
2. The plane containing the lines \(\frac{x}{3} = \frac{y}{4} = \frac{z}{2}\) and \(\frac{x}{4} = \frac{y}{2} = \frac{z}{3}\) will have a normal vector along the cross product of their direction vectors:
   • Direction vector of line 1: \(\langle 3, 4, 2 \rangle\)
   • Direction vector of line 2: \(\langle 4, 2, 3 \rangle\)
3. Compute the cross product: \(\langle 3, 4, 2 \rangle \times \langle 4, 2, 3 \rangle = \langle (4 \times 3 - 2 \times 2), (2 \times 4 - 3 \times 3), (3 \times 2 - 4 \times 4) \rangle = \langle 8, -1, -10 \rangle\).
4. The plane we find needs to be perpendicular to the vector \(\langle 8, -1, -10 \rangle\), which will serve as the normal vector for our desired plane. Additionally, this plane contains the direction ratios 2, 3, 4 of the line \(\frac{x}{2} = \frac{y}{3} =\frac{z}{4}\).
5. The equation of the plane can be expressed in terms of general form using the normal vector: \(ax + by + cz = 0\). Substituting the given direction ratios of line, \(a(2) + b(3) + c(4) = 0\). Considering the coefficients are proportional to the normal vector to both conditions: normal and containing the line, we apply the known normal.
6. Test the options to find the suitable plane:
   • For \(x - 2y + z = 0\), substituting a point from the line might not be immediate. Check if direction (resultant plane normal should be orthogonal to the plane): \(8 \times 1 + (-1) \times (-2) + (-10) \times 1 = 8 + 2 - 10 = 0\), indicating orthogonality.

Thus, the equation of the plane satisfying both conditions is \(x - 2y + z = 0\).