Consider two lines l_1 and l_2 with cartesian equations x/2 = (1-y)/(-2) = z/1 and (2x-5)/16 = (y-2)/(-1) = (z-5)/4 respectively. Which of the following is/are true? (A) Direction ratio of l_1 are 2, 2, 1 (B) Direction cosines of l_1 are 2/3, (-2)/3, 1/3 (C) Direction ratio of l_2 are 16, -1, 4 (D) Angle between l_1 and l_2 is ^(-1)(38/3√273) Choose the correct answer from the options given below: Option 1: (B), (C) and (D) only Option 2: (A) and (B) only Option 3: (C) and (D) only Option 4: (A) only

Answer: (A) only. Solution: The standard form of a line in 3D is: (x - x_1)/a = (y - y_1)/b = (z - z_1)/c where (a, b, c) are the direction ratios. l_1: x/2 = (1-y)/(-2) = z/1 The y-part needs to be converted to standard form: (1-y)/(-2) = (-(y-1))/(-2) = (y-1)/2 So l_1 in standard form is: (x-0)/2 = (y-1)/2 = (z-0)/1 Direction ratios of l_1 = (2,\ 2,\ 1) (A) is TRUE, DR, = √(2^2 + 2^2 + 1^2) = √9 = 3 Direction cosines = (2/3,\ 2/3,\ 1/3) Statement (B) claims DCs are (2/3,\ (-2)/3,\ 1/3) — the second component should be positive, not negative. (B) is FALSE l_2: (2x-5)/16 = (y-2)/(-1) = (z-5)/4 The x-part has a coefficient of 2 in front of x, so factoring it out: (2x - 5)/16 = (2(x - 5/2))/16 = (x - 5/2)/8 So l_2 in standard form is: (x - 5/2)/8 = (y-2)/(-1) = (z-5)/4 Direction ratios of l_2 = (8,\ -1,\ 4), not (16,\ -1,\ 4) (C) is FALSE Using the correct direction ratios: l_1: (2,\ 2,\ 1) l_2: (8,\ -1,\ 4) = (, b_1 * b_2, )/(, b_1, \, b_2, ) = (, 2(8) + 2(-1) + 1(4), )/(√(4+4+1) * √(64+1+16)) = (, 16 - 2 + 4, )/(3 x √81) = 18/(3 x 9) = 2/3 = ^(-1)(2/3) Statement (D) claims = ^(-1)(38/3√273), which results from using incorrect direction ratios. (D) is FALSE Only statement (A) is true. The correct answer is (A) only.

Geometry

3D Geometry

Hard

core

Consider two lines $l_1$ and $l_2$ with cartesian equations $\frac{x}{2} = \frac{1-y}{-2} = \frac{z}{1}$ and $\frac{2x-5}{16} = \frac{y-2}{-1} = \frac{z-5}{4}$ respectively. Which of the following is/are true?

(A) Direction ratio of $l_1$ are 2, 2, 1

(B) Direction cosines of $l_1$ are $\frac{2}{3}, \frac{-2}{3}, \frac{1}{3}$

(C) Direction ratio of $l_2$ are 16, -1, 4

(D) Angle between $l_1$ and $l_2$ is $\cos^{-1}\left(\frac{38}{3\sqrt{273}}\right)$

Choose the correct answer from the options given below:

(B), (C) and (D) only

(A) and (B) only

(A) only

✅ Correct Option: 4

The standard form of a line in 3D is:

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$

where $(a, b, c)$ are the direction ratios.

$l_1: \quad \frac{x}{2} = \frac{1-y}{-2} = \frac{z}{1}$

The $y$ -part needs to be converted to standard form:

$\frac{1-y}{-2} = \frac{-(y-1)}{-2} = \frac{y-1}{2}$

So $l_1$ in standard form is:

$\frac{x-0}{2} = \frac{y-1}{2} = \frac{z-0}{1}$

Direction ratios of $l_1 = (2,\ 2,\ 1)$

(A) is TRUE

$|\text{DR}| = \sqrt{2^2 + 2^2 + 1^2}$

$= \sqrt{9} = 3$

Direction cosines $= \left(\frac{2}{3},\ \frac{2}{3},\ \frac{1}{3}\right)$

Statement (B) claims DCs are $\left(\frac{2}{3},\ \frac{-2}{3},\ \frac{1}{3}\right)$ — the second component should be positive, not negative.

(B) is FALSE

$l_2: \quad \frac{2x-5}{16} = \frac{y-2}{-1} = \frac{z-5}{4}$

The $x$ -part has a coefficient of $2$ in front of $x$ , so factoring it out:

$\frac{2x - 5}{16} = \frac{2\left(x - \frac{5}{2}\right)}{16} = \frac{x - \frac{5}{2}}{8}$

So $l_2$ in standard form is:

$\frac{x - \frac{5}{2}}{8} = \frac{y-2}{-1} = \frac{z-5}{4}$

Direction ratios of $l_2 = (8,\ -1,\ 4)$ , not $(16,\ -1,\ 4)$

Using the correct direction ratios:

$l_1$ : $(2,\ 2,\ 1)$

$l_2$ : $(8,\ -1,\ 4)$

$\cos\theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}|\ |\vec{b_2}|}$

$= \frac{|2(8) + 2(-1) + 1(4)|}{\sqrt{4+4+1} \cdot \sqrt{64+1+16}}$

$= \frac{|16 - 2 + 4|}{3 \times \sqrt{81}}$

$= \frac{18}{3 \times 9}$

$= \frac{2}{3}$

$\theta = \cos^{-1}\left(\frac{2}{3}\right)$

Statement (D) claims $\theta = \cos^{-1}\left(\frac{38}{3\sqrt{273}}\right)$ , which results from using incorrect direction ratios.

(D) is FALSE

Only statement (A) is true.

The correct answer is (A) only.

2023: 22 May Shift 3

Distance between the point (3, 4, 5) and the point where the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ meets the plane $x + y + z = 17$ is :

2023: 22 May Shift 3

Cartesian equation of plane passing through the points (2, -4, 5) and perpendicular to the line with direction ratios (3, -1, 2) is :

2022: 25 May Shift 1

A. Equation of the line passing through the point (1, 2, 3) and parallel to the vector $3\hat{i} + 2\hat{j} - 2\hat{k}$ is $\frac{x-1}{3} = \frac{y-2}{2} = \frac{y-3}{-2}$ .

B. Equation of line passing through (1, 2, 3) and parallel to the line given by $\frac{x+3}{3} = \frac{4-y}{5} = \frac{z+8}{6}$ is $\frac{x-1}{3} = \frac{y-2}{5} = \frac{z+3}{6}$ .

C. Equation of line passing through the origin and (5, -2, 3) is $\frac{x}{5} = \frac{y}{-2} = \frac{z}{3}$ .

D. Equation of plane passing through the point (1, 2, 3) and perpendicular to the line with direction ratio's 2, 3, -1 is $2(x-1)+3(y-2)-1(z-3) = 0$ .

E. Equation of plane with intercepts 2, 3 and 4 on x, y and z-axis respectively is $2x + 3y + 4z = 1$ .

Choose the correct answer from the options given below:

2026: 21 May Shift 2

Let the square of the shortest distance between the lines $L_1: \vec{r} = (\hat{i}+2\hat{j}-4\hat{k}) + \lambda(2\hat{i}+3\hat{j}+6\hat{k})$ and $L_2: \vec{r} = (3\hat{i}+3\hat{j}-5\hat{k}) + \mu(2\hat{i}+3\hat{j}+6\hat{k})$ is $\dfrac{m}{n}$ , where $\gcd(m,n)=1$ . Then $m+n$ is equal to:

2026: 22 May Shift 2

Match List-I with List-II

List-I	List-II
(A) The distance of a point $P(a,b,c)$ from x-axis	(I) $\sqrt{a^2+b^2+c^2}$
(B) The distance of a point $P(a,b,c)$ from y-axis	(II) $\sqrt{b^2+c^2}$
(C) The distance of a point $P(a,b,c)$ from Z-axis	(III) $\sqrt{a^2+c^2}$
(D) The distance of a point $P(a,b,c)$ from origin	(IV) $\sqrt{a^2+b^2}$

Choose the correct answer from the options given below:

2026: 30 May Shift 1

Let $L_1$ and $L_2$ lines such that $L_1:\vec{r}=\hat{i}+\hat{j}+\lambda(2\hat{i}-\hat{j}+\hat{k})$ , $L_2:\vec{r}=2\hat{i}+\hat{j}-\hat{k}+\mu(4\hat{i}-2\hat{j}+2\hat{k})$ and $\theta$ be the acute angle between $L_1$ and $L_2$ , then

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	$\cos\theta$	I.	$\dfrac{\sqrt{66}}{6}$
B.	$\sin\theta$	II.	$3$
C.	the shortest distance (in units) between $L_1$ and $L_2$	III.	$1$
D.	(sum of squares of direction cosines of $L_1$ ) $+2$	IV.	$0$

Choose the correct answer from the options given below:

Solution

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