Math Problem Statement

If the straight lines are coplanar find the real values of m. (x - 1)/1 = (y - 2)/2 = (z - 3)/(m ^ 2) (x - 3)/1 = (y - 2)/(m ^ 2) = (z - 1)/2

Solution

To find the real values of mm such that the given straight lines are coplanar, we need to use the condition for coplanarity of two lines. The lines in symmetric form are given by:

  1. x11=y22=z3m2\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{m^2}
  2. x31=y2m2=z12\frac{x - 3}{1} = \frac{y - 2}{m^2} = \frac{z - 1}{2}

The condition for the coplanarity of two lines given in symmetric form:

Lines xx1a1=yy1b1=zz1c1 and xx2a2=yy2b2=zz2c2 are coplanar if the determinant:\text{Lines } \frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1} \text{ and } \frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2} \text{ are coplanar if the determinant:}

x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$$ ### Step 1: Identify the parameters For the first line: - Point $$P_1$$ is $$(1, 2, 3)$$ - Direction ratios: $$a_1 = 1$$, $$b_1 = 2$$, $$c_1 = m^2$$ For the second line: - Point $$P_2$$ is $$(3, 2, 1)$$ - Direction ratios: $$a_2 = 1$$, $$b_2 = m^2$$, $$c_2 = 2$$ ### Step 2: Apply the coplanarity condition The vector joining the points $$P_1$$ and $$P_2$$ is: $$\overrightarrow{P_1P_2} = (3 - 1, 2 - 2, 1 - 3) = (2, 0, -2)$$ Now, apply the coplanarity condition: $$\begin{vmatrix} 2 & 0 & -2 \\ 1 & 2 & m^2 \\ 1 & m^2 & 2 \end{vmatrix} = 0$$ ### Step 3: Calculate the determinant Expand the determinant: $$= 2 \left(2 \cdot 2 - m^2 \cdot m^2\right) - 0 \cdot (1 \cdot 2 - m^2 \cdot 1) - 2 \left(1 \cdot m^2 - 2 \cdot 1\right)$$ $$= 2 \left(4 - m^4\right) - 2 \left(m^2 - 2\right)$$ $$= 8 - 2m^4 - 2m^2 + 4$$ $$= -2m^4 - 2m^2 + 12$$ Set the determinant to zero: $$-2m^4 - 2m^2 + 12 = 0$$ ### Step 4: Solve the equation Divide the entire equation by $$-2$$: $$m^4 + m^2 - 6 = 0$$ Let $$t = m^2$$. The equation becomes: $$t^2 + t - 6 = 0$$ Factorizing the quadratic: $$(t - 2)(t + 3) = 0$$ So, $$t = 2$$ or $$t = -3$$. Since $$t = m^2$$, and $$m^2$$ cannot be negative: $$m^2 = 2 \Rightarrow m = \pm \sqrt{2}$$ ### Final Answer: The real values of $$m$$ for which the lines are coplanar are $$m = \sqrt{2}$$ and $$m = -\sqrt{2}$$. Would you like a detailed explanation of any step? Here are 5 related questions: 1. How do you determine the direction ratios of a line given in symmetric form? 2. What is the general condition for two lines to be coplanar? 3. Can we extend this method to 3D planes intersecting lines? 4. How do we solve higher-degree polynomials like quartic equations? 5. What are some applications of coplanarity in geometry? **Tip:** When working with determinants, always simplify the matrix first to avoid errors in expansion.

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Math Problem Analysis

Mathematical Concepts

Vectors
Systems of Equations
Determinants

Formulas

Equation of a line in symmetric form
Condition for coplanarity of lines
Determinant formula for coplanarity

Theorems

Coplanarity theorem for lines

Suitable Grade Level

Advanced High School