Let the values of p, for which the shortest distance between the lines (x+1)/3 = y/4 = z/5 and r = (pi + 2j + k) + lambda(2i + 3j + 4k) is 1/sqrt(6), be a, b, (a < b). Then the length of the latus rectum of the ellipse x^2/a^2 + y^2/b^2 = 1 is :

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q14 – JEE-MAIN 2025

← Back

Home Maths 3D GeometryQuestion #25

Let the values of p, for which the shortest distance between the lines

\frac{x + 1}{3} = \frac{y}{4} = \frac{z}{5}

and

\vec{r} = (p \hat{i} + 2 \hat{j} + \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k})

\frac{1}{\sqrt{6}}

, be a, b, (a < b). Then the length of the latus rectum of the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

is :

Clue 1 of 3: Shortest Distance between Skew Lines

Recall the formula for the shortest distance between two skew lines given in vector form.

🧠 Full Solution Path

Step 1: Identify Line Parameters and Calculate Shortest Distance✦ Active

The first line is $\frac{x - (- 1)}{3} = \frac{y - 0}{4} = \frac{z - 0}{5}$ , so $\vec{a_{1}} = - \hat{i}$ and $\vec{b_{1}} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}$ . The second line is $\vec{r} = (p \hat{i} + 2 \hat{j} + \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k})$ , so $\vec{a_{2}} = p \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{b_{2}} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ . Calculate the necessary vector products:

\vec{a_{2}} - \vec{a_{1}} = (p + 1) \hat{i} + 2 \hat{j} + \hat{k}

\vec{b_{1}} \times \vec{b_{2}} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{matrix} | = \hat{i} (16 - 15) - \hat{j} (12 - 10) + \hat{k} (9 - 8) = \hat{i} - 2 \hat{j} + \hat{k}

| \vec{b_{1}} \times \vec{b_{2}} | = \sqrt{1^{2} + (- 2)^{2} + 1^{2}} = \sqrt{6}

(\vec{a_{2}} - \vec{a_{1}}) \cdot (\vec{b_{1}} \times \vec{b_{2}}) = (p + 1) (1) + (2) (- 2) + (1) (1) = p + 1 - 4 + 1 = p - 2

The shortest distance $d = \frac{| p - 2 |}{\sqrt{6}}$ . Given $d = \frac{1}{\sqrt{6}}$ , we have $| p - 2 | = 1$ , which yields $p - 2 = 1$ or $p - 2 = - 1$ . Thus, $p = 3$ or $p = 1$ .

Step 2: Determine 'a' and 'b' and Ellipse Parameters○ Expand

The values of $p$ are $1$ and $3$ . Since $a < b$ , we have $a = 1$ and $b = 3$ . The ellipse equation is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , which becomes $\frac{x^{2}}{1^{2}} + \frac{y^{2}}{3^{2}} = 1$ , or $\frac{x^{2}}{1} + \frac{y^{2}}{9} = 1$ . Comparing this to the standard form $\frac{x^{2}}{A^{2}} + \frac{y^{2}}{B^{2}} = 1$ , we have $A^{2} = 1$ and $B^{2} = 9$ . Since $B^{2} > A^{2}$ , the major axis is along the y-axis, with $A = 1$ and $B = 3$ .

💡 Teacher's Secret Hint

Remember to correctly identify the major and minor axes based on the values of $a^{2}$ and $b^{2}$ in the ellipse equation.

Step 3: Calculate Length of Latus Rectum○ Expand

For an ellipse with the major axis along the y-axis, the length of the latus rectum is given by $\frac{2 A^{2}}{B}$ . Substituting the values $A = 1$ and $B = 3$ :

Length of latus rectum = \frac{2 (1)^{2}}{3} = \frac{2}{3}

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 14 – JEE-MAIN 2025

Clue 1 of 3: Shortest Distance between Skew Lines

🧠 Full Solution Path