If the shortest distance between the lines l_1 and l_2 given by r = ai + 2j - k + (2i - j + k) and r = i - j + k + (2i - j + k) is √(35/6) units, the values of 'a' can be : Option 1: 0, -8 Option 2: 0, 8 Option 3: 2, 6 Option 4: -2, 6

Answer: 0, -8. Solution: Both lines have direction b=(2,-1,1), so they are parallel. a_2-a_1=(1-a, -3, 2). (a_2-a_1)xb=(-1, 3+a, 5+a). Distance: √((1+(3+a)^2+(5+a)^2)/6)=√(35/6). Solving: 2a^2+16a=0 a=0 or a=-8.

Geometry

3D Geometry

Hard

core

If the shortest distance between the lines $l_1$ and $l_2$ given by $\vec{r} = a\hat{i} + 2\hat{j} - \hat{k} + \lambda(2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = \hat{i} - \hat{j} + \hat{k} + \mu(2\hat{i} - \hat{j} + \hat{k})$ is $\sqrt{\frac{35}{6}}$ units, the values of 'a' can be :

0, -8

0, 8

2, 6

-2, 6

✅ Correct Option: 1

Both lines have direction $\vec{b}=(2,-1,1)$ , so they are parallel.

$\vec{a_2}-\vec{a_1}=(1-a, -3, 2)$ .

$(\vec{a_2}-\vec{a_1})\times\vec{b}=(-1, 3+a, 5+a)$ .

Distance: $\sqrt{\frac{1+(3+a)^2+(5+a)^2}{6}}=\sqrt{\frac{35}{6}}$ .

Solving: $2a^2+16a=0 \Rightarrow a=0$ or $a=-8$ .

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: 23 May Shift 3

The value of $\lambda$ , so that the lines $\frac{1-x}{3} = \frac{7y-14}{2\lambda} = \frac{z-3}{2}$ and $\frac{7-7x}{3\lambda} = \frac{y-5}{1} = \frac{6-z}{5}$ are perpendicular is:

2025: 2 June Shift 1

If the direction ratios of two lines are $a, b, c$ and $(b-c), (c-a), (a-b)$ respectively, then the angle between these lines is:

2026: 29 May Shift 2

If $\lambda$ and $\mu$ are parameters, then

Match the LIST-I with LIST-II

LIST-I		LIST-II
A.	$\vec{r}=2\hat{i}+\hat{j}-3\hat{k}+\lambda(\hat{i}-\hat{j}+2\hat{k})$ parallel to the line	I. $\vec{r}=(\hat{j}+\hat{k})+\mu(-2\hat{i}+\hat{j}-2\hat{k})$
B.	$\vec{r}=\hat{i}+\hat{j}+\lambda(\hat{i}+\hat{j}-2\hat{k})$ perpendicular to the line	II. $\vec{r}=(2\hat{i}+\hat{k})+\mu(-2\hat{i}+\hat{j})$
C.	$\frac{x}{-2}=\frac{y-1}{1}=\frac{z-1}{-2}$ in vector form	III. $\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu(2\hat{i}-2\hat{j}+4\hat{k})$
D.	$\frac{2-x}{2}=y, z=1$ in vector form	IV. $\frac{x-1}{2}=\frac{y+1}{4}=\frac{z}{3}$

Choose the correct answer from the options given below:

2023: 15 June Shift 2

If the straight lines $x = 1 + s$ , $y = -3 - \lambda s$ , $z = 1 + \lambda s$ and $x = \frac{t}{2}$ , $y = 1 + t$ , $z = 2 - t$ with parameters $s$ and $t$ respectively, are coplanar, then $\lambda$ is equal to :

2023: 15 June Shift 2

The equation of plane which cuts equal intercepts of unit length on the coordinate axes is :

Solution

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