List of top Mathematics Questions on Three Dimensional Geometry asked in MHT CET

A line L is passing through points A(1, 3, 2) and B(2, 2, 1). If mirror image of point P(1, 1, -1) in the line L is (x, y, z) then $x + y + z =$

The co-ordinates of the point in which line joining $(1, 1, 1)$ and $(2, 2, 2)$ intersects the plane $x + y + z = 9$ are

The value of $\int_{-3}^{3} \sin^7 x \cos^{16} x \, dx$ is

If $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$ and $A^{-1} = \alpha I + \beta A$, $\beta \in R$ where I is the identity matrix of order 2, then $4(\alpha + \beta) =$}

The value of \( \int_{1/3}^{1} \frac{(x - x^3)^{\frac{1{3}}}{x^4} dx \) is

If \( x^{\frac{2}{5}} + y^{\frac{2}{5}} = \text{a}^{\frac{2}{5}} \) then \( \frac{dy}{dx} = \)

If \( X \sim B(n, p) \) then \( \frac{P(X=k){P(X=k-1)} = \)}

The number of solutions of $\tan^{-1} (x + \frac{2}{x}) - \tan^{-1} (\frac{4}{x}) - \tan^{-1} (x - \frac{2}{x}) = 0$ are

Derivative of $x^{(x^x)}$ is

The correct constraints for the given feasible region are ..............

If the point $(1, \alpha, \beta)$ lies on the line of the shortest distance between the lines $\frac{x+2}{-3} = \frac{y-2}{4} = \frac{z-5}{2}$ and $\frac{x+2}{-1} = \frac{y+6}{2}, z = 1$, then $\alpha + \beta =$

If $\frac{1}{6} \sin \theta, \cos \theta, \tan \theta$ are in G.P., then the general solution of $\theta$ is

The lines $x + 2ay + \text{a} = 0, x + 3by + \text{b} = 0, x + 4cy + \text{c} = 0$ are concurrent then $a, b, c$ are in}

A line L is passing through points A(1, 3, 2) and B(2, 2, 1). If mirror image of point P(1, 1, -1) in the line L is (x, y, z) then $x + y + z =$

The co-ordinates of the point in which line joining $(1, 1, 1)$ and $(2, 2, 2)$ intersects the plane $x + y + z = 9$ are

The angle between the lines, whose direction cosines \( l, m, n \) satisfy the equations: \[ l + m + n = 0 \quad \text{and} \quad 2l^2 + 2m^2 - n^2 = 0, \] is:

The distance of the point having position vector $\hat{i}-2\hat{j}-6\hat{k}$ from the straight line passing through the point $(2,-3,-4)$ and parallel to the vector $6\hat{i}+3\hat{j}-4\hat{k}$ is

The length of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$ is

A vector parallel to the line of intersection of the planes $\vec{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1$ and $\vec{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2$ is

The mirror image of $P(2, 4, -1)$ in the plane $x - y + 2z - 2 = 0$ is $(a, b, c)$, then the value of $a+b+c$ is

Two lines \(2x=3y=-z\) and \(6z=-y=-4z\) intersect at a point, then find the angle between them.