Distance between two planes x + 2y - z = 5 and 2x + 4y - 2z + 2 = 0 is Option 1: √6 unit Option 2: 7 unit Option 3: 5/√6 unit Option 4: 4/√6 unit

Answer: √6 unit. Solution: Make coefficients identical. First plane x 2: 2x+4y-2z = 10. Second plane: 2x+4y-2z = -2. Distance = (, 10-(-2), )/(√(2^2+4^2+(-2)^2)) = 12/√24 = 12/2√6 = √6 units.

Geometry

3D Geometry

Easy

Distance between two planes $x + 2y - z = 5$ and $2x + 4y - 2z + 2 = 0$ is

$\sqrt{6}$ unit

$7$ unit

$\frac{5}{\sqrt{6}}$ unit

$\frac{4}{\sqrt{6}}$ unit

✅ Correct Option: 1

Make coefficients identical. First plane $\times 2$ : $2x+4y-2z = 10$ . Second plane: $2x+4y-2z = -2$ .

Distance $= \frac{|10-(-2)|}{\sqrt{2^2+4^2+(-2)^2}} = \frac{12}{\sqrt{24}} = \frac{12}{2\sqrt{6}} = \sqrt{6}$ units.

2026: 19 May Shift 1

If $\theta$ is the angle between the lines $L_1$ and $L_2$ , whose direction cosines are $l_1, m_1, n_1$ and $l_2, m_2, n_2$ respectively, then

Match List-I with List-II

List-I	List-II
(A) $\cos\theta$	(I) $l_1l_2+m_1m_2+n_1n_2$
(B) $\sin\theta$	(II) $l_1l_2+m_1m_2+n_1n_2=0$
(C) $L_1$ and $L_2$ are perpendicular	(III) $\frac{l_1}{l_2}=\frac{m_1}{m_2}=\frac{n_1}{n_2}$
(D) $L_1$ and $L_2$ are parallel	(IV) $\sqrt{\sum(m_1n_2-m_2n_1)^2}$

Choose the correct answer from the options given below:

2026: 21 May Shift 2

Let ABCD be a parallelogram with vertices $A(2,2,a), B(1,0,0), C(1,3,a)$ and D, where $a > 0$ . If the area of parallelogram is 5 square units, then the value of $a$ is:

2023: 23 May Shift 3

Match List I with List II

LIST I	LIST II
A. $lx + my + nz = d$ is	I. Equation of plane passing through a given point and normal to given vector
B. $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ is	II. Equation of plane in normal form
C. $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$	III. Plane passing through the intersection of two planes
D. $(a_1x + b_1y + c_1z + d_1) + \lambda(a_2x + b_2y + c_2z + d_2) = 0$	IV. Intercept from of plane

Choose the correct answer from the options given below:

2025: 15 May Shift 1

Consider the lines $l_1: \frac{x-1}{0} = \frac{y-1}{1} = \frac{2-z}{1}$ and $l_2: \frac{x}{2} = \frac{y}{0} = \frac{2z-1}{4}$ , then which of the following are correct?

(A) Direction Ratio's of $l_1 = <0, 1, 1>$

(B) Direction Ratio's of $l_2 = <2, 0, 2>$

(D) Angle between $l_1$ and $l_2 =$ $\frac{2\pi}{3}$

Choose the correct answer from the options given below:

2026: 26 May Shift 2

If the points $P(3,\mu,6), Q(-1,-1,2)$ and $R(2,8,5)$ are collinear, then the value of ' $\mu$ ' is

2026: 15 May Shift 1

Match the LIST-I with LIST-II

	LIST-I (Equation of Line)		LIST-II (Direction Ratio of line)
A.	$\dfrac{2-x}{1} = \dfrac{2y+1}{1} = \dfrac{z}{1}$	I.	$2, 1, 1$
B.	$x = 2y+3,\ z = y+1$	II.	$1, -2, 2$
C.	$\dfrac{x-1}{1} = \dfrac{y+1}{2},\ z = 1$	III.	$-2, 1, 2$
D.	$\vec{r} = \hat{i} + \hat{j} + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})$	IV.	$1, 2, 0$

Choose the correct answer from the options given below:

Distance between two planes x+2y−z=5x + 2y - z = 5x+2y−z=5 and 2x+4y−2z+2=02x + 4y - 2z + 2 = 02x+4y−2z+2=0 is

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

Distance between two planes $x + 2y - z = 5$ and $2x + 4y - 2z + 2 = 0$ is