Let l1 be the line in xy-plane with x and y intercepts ⅛ and \(\frac{1}{4√2} \) respectively and l2 be the line in zx-plane with x and z intercepts -⅛ and \(−\frac{1}{6√3}\) respectively. If d is the shortest distance between the line l1 and l2, then d–2 is equal to ______.
Correct Answer: 51
Solution and Explanation
The correct answer is: 51.
\(\frac{x-\frac{1}{8}}{\frac{1}{8}}=\frac{y}{-\frac{1}{4\sqrt2}}=\frac{z}{0}....L_1\)
or
\(\frac{x-\frac{1}{8}}{1}=\frac{y}{-\sqrt2}=\frac{z}{0}....(i)\)
\(=\frac{\sqrt2}{\sqrt{32+16+54}}=\frac{!}{\sqrt51}\)
d–2 = 51
Top Questions on Plane
- Let the foot of perpendicular from a point \( P(1,2,-1) \) to the straight line \( L : \frac{x}{1} = \frac{y}{0} = \frac{z}{-1} \) be \( N \). Let a line be drawn from \( P \) parallel to the plane \( x + y + 2z = 0 \) which meets \( L \) at point \( Q \). If \( \alpha \) is the acute angle between the lines \( PN \) and \( PQ \), then \( \cos \alpha \) is equal to:
- Let R be the relation "is congruent to" on the set of all triangles in a plane. Is R:
- Let the acute angle bisector of the two planes \( x - 2y - 2z + 1 = 0 \) and \( 2x - 3y - 6z + 1 = 0 \) be the plane \( P \). Then which of the following points lies on \( P \)?
- If the plane \[ x - y + z + 4 = 0 \] divides the line joining the points \[ P(2,3,-1) \quad {and} \quad Q(1,4,-2) \] in the ratio \( l:m \), then \( l + m \) is:
- The equation of the plane passing through the point \( (1, 1, 1) \) and perpendicular to the planes \( 2x + y - 2z = 5 \) and \( 3x - 6y - 2z = 7 \) is:
Questions Asked in JEE Main exam
- Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is:
- JEE Main - 2025
- Solutions
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.- JEE Main - 2025
- Electrical Circuits
- Let \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k} \) and \( \vec{b} = 2\hat{i} + 7\hat{j} + 3\hat{k} \). Let \( L_1 : \vec{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda \vec{a}, \lambda \in {R} \) and \( L_2 : \vec{r} = (\hat{j} + \hat{k}) + \mu \vec{b}, \mu \in \mathbb{R} \) be two lines. If the line \( L_3 \) passes through the point of intersection of \( L_1 \) and \( L_2 \), and is parallel to \( \vec{a} + \vec{b} \), then \( L_3 \) passes through the point:
- JEE Main - 2025
- Vectors
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
- JEE Main - 2025
- Relations and functions
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
Concepts Used:
Distance of a Point from a Plane
The shortest perpendicular distance from the point to the given plane is the distance between point and plane. In simple terms, the shortest distance from a point to a plane is the length of the perpendicular parallel to the normal vector dropped from the particular point to the particular plane. Let's see the formula for the distance between point and plane.
Read More: Distance Between Two Points