The distance of the point (1, 2, 0) from the line x-3/2 = y+4/3 = z+6/5 measured parallel to the plane x + y + z = 3 is Option 1: sqrt(26) Option 2: sqrt(89) Option 3: 0 Option 4: sqrt(27)

The answer is 'sqrt(26)'. View full solution:

CUET Mathematics - The distance of the point (1, 2, 0) from the line x-3/2 = y+4/3 = z+6/5 measured parallel to the plane x + y + z = 3 is | PYQs + Solutions

CUET Mathematics 2022 4 Aug Shift 1

Geometry

Hard

$\sqrt{26}$

$\sqrt{89}$

$\sqrt{27}$

✅ Correct Option: 1

Related questions:

22 May Shift 1

Match List-I with List-II

List-I	List-II
(A) Equations of line through $(5, -4, 6)$ with direction ratios $3, 7, 2$	(I) $\frac{x+3}{5} = \frac{y+7}{-4} = \frac{z+2}{6}$
(B) Equations of line through $(3, 7, 2)$ with direction ratios $5, -4, 6$	(II) $\frac{x-3}{5} = \frac{y-7}{-4} = \frac{z-2}{6}$
(C) Equations of line through $(-5, 4, -6)$ with direction ratios $3, 7, 2$	(III) $\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}$
(D) Equations of line through $(-3, -7, -2)$ with direction ratios $5, -4, 6$	(IV) $\frac{x+5}{3} = \frac{y-4}{7} = \frac{z+6}{2}$

Choose the correct answer from the options given below:

15 May Shift 1

The straight line $\frac{x+3}{3} = \frac{y+2}{4} = \frac{z+1}{0}$ is

13 May Shift 2

The shortest distance between the lines $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$ and $\vec{r} = (2\hat{i} + 4\hat{j} + 5\hat{k}) + \mu(3\hat{i} + 4\hat{j} + 5\hat{k})$ is equal to

CUET Mathematics 2022 4 Aug Shift 1 PYQs

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