Q1366: Verify that lines given by $\vec{r}=(1-\lambda)\hat{i}+(\lam...

Class CBSE Class 12 Mathematics Three Dimensional Geometry Q #1366

COMPETENCY BASED

UNDERSTAND

3 Marks 2025 AISSCE(Board Exam) SA

Verify that lines given by $\vec{r}=(1-\lambda)\hat{i}+(\lambda-2)\hat{j}+(3-2\lambda)\hat{k}$ and $\vec{r}=(\mu+1)\hat{i}+(2\mu-1)\hat{j}-(2\mu+1)\hat{k}$ are skew lines. Hence, find shortest distance between the lines.

Prev Next

AI Tutor Explanation

Detailed Solution

Step 1: Rewrite the equations in standard form

The given equations are: $\vec{r}=(1-\lambda)\hat{i}+(\lambda-2)\hat{j}+(3-2\lambda)\hat{k}$ $\vec{r}=(\mu+1)\hat{i}+(2\mu-1)\hat{j}-(2\mu+1)\hat{k}$ We can rewrite these as: $\vec{r}=(\hat{i}-2\hat{j}+3\hat{k}) + \lambda(-\hat{i}+\hat{j}-2\hat{k})$ $\vec{r}=(\hat{i}-\hat{j}-\hat{k}) + \mu(\hat{i}+2\hat{j}-2\hat{k})$ Comparing with the standard form $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$, we have: $\vec{a_1} = \hat{i}-2\hat{j}+3\hat{k}$ $\vec{b_1} = -\hat{i}+\hat{j}-2\hat{k}$ $\vec{a_2} = \hat{i}-\hat{j}-\hat{k}$ $\vec{b_2} = \hat{i}+2\hat{j}-2\hat{k}$

Step 2: Check if the lines are parallel

For the lines to be parallel, $\vec{b_1}$ and $\vec{b_2}$ must be proportional. $\frac{-1}{1} \neq \frac{1}{2} \neq \frac{-2}{-2}$ Since the ratios are not equal, the lines are not parallel.

Step 3: Check if the lines intersect

For the lines to intersect, the shortest distance between them must be zero. The shortest distance is given by: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$ If $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0$, the lines intersect. $\vec{a_2} - \vec{a_1} = (\hat{i}-\hat{j}-\hat{k}) - (\hat{i}-2\hat{j}+3\hat{k}) = 0\hat{i} + \hat{j} - 4\hat{k}$ $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(-2+4) - \hat{j}(2+2) + \hat{k}(-2-1) = 2\hat{i} - 4\hat{j} - 3\hat{k}$ $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (0\hat{i} + \hat{j} - 4\hat{k}) \cdot (2\hat{i} - 4\hat{j} - 3\hat{k}) = 0 - 4 + 12 = 8$ Since $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \neq 0$, the lines do not intersect.

Step 4: Conclude that the lines are skew

Since the lines are neither parallel nor intersecting, they are skew lines.

Step 5: Calculate the shortest distance

The shortest distance between the skew lines is given by: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$ We already have $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 8$ $|\vec{b_1} \times \vec{b_2}| = \sqrt{2^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$ Therefore, the shortest distance is: $d = \left| \frac{8}{\sqrt{29}} \right| = \frac{8}{\sqrt{29}}$

Step 6: Rationalize the denominator

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{29}$: $d = \frac{8}{\sqrt{29}} \times \frac{\sqrt{29}}{\sqrt{29}} = \frac{8\sqrt{29}}{29}$

Final Answer: $\frac{8\sqrt{29}}{29}$

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit

Bloom's Analysis: This is an APPLY question because it requires applying the formulas and concepts of vectors and 3D geometry to determine if lines are skew and to calculate the shortest distance between them.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to follow a specific procedure to verify if the lines are skew and then apply the formula for the shortest distance between two skew lines.

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It assesses the student's ability to apply vector algebra concepts to solve problems related to 3D geometry, specifically identifying skew lines and calculating shortest distances.