Cosine of the acute angle between the lines (x-3)/2 = (y-2)/1 = (z-5)/2 and (x-1)/6 = (y-3)/(-3) = (z+6)/2 is Option 1: 11/21 Option 2: 13/21 Option 3: 10/21 Option 4: 8/21

Answer: 13/21. Solution: The angle between two lines is the angle between their direction vectors. For the first line, the direction vector is v_1 = 2i + 1j + 2k, and for the second line, it is v_2 = 6i - 3j + 2k. The cosine of the angle is given by the formula = (, v_1 * v_2, )/(, v_1, v_2, ). Calculating the dot product gives (2)(6) + (1)(-3) + (2)(2) = 12 - 3 + 4 = 13. The magnitude of the first vector is √(2^2 + 1^2 + 2^2) = √9 = 3. The magnitude of the second vector is √(6^2 + (-3)^2 + 2^2) = √(36 + 9 + 4) = √49 = 7. Substituting these into the formula, we get = 13/(3 x 7) = 13/21

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Cosine of the acute angle between the lines $\frac{x-3}{2} = \frac{y-2}{1} = \frac{z-5}{2}$ and $\frac{x-1}{6} = \frac{y-3}{-3} = \frac{z+6}{2}$ is

$\frac{11}{21}$

$\frac{13}{21}$

$\frac{10}{21}$

$\frac{8}{21}$

✅ Correct Option: 2

The angle between two lines is the angle between their direction vectors.

For the first line, the direction vector is $\vec{v}_1 = 2\hat{i} + 1\hat{j} + 2\hat{k}$ , and for the second line, it is $\vec{v}_2 = 6\hat{i} - 3\hat{j} + 2\hat{k}$ .

The cosine of the angle $\theta$ is given by the formula $\cos \theta = \frac{|\vec{v}_1 \cdot \vec{v}_2|}{|\vec{v}_1| |\vec{v}_2|}$ .

Calculating the dot product gives $(2)(6) + (1)(-3) + (2)(2) = 12 - 3 + 4 = 13$ .

The magnitude of the first vector is $\sqrt{2^2 + 1^2 + 2^2} = \sqrt{9} = 3$ .

The magnitude of the second vector is $\sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$ .

Substituting these into the formula, we get $\cos \theta = \frac{13}{3 \times 7} = \frac{13}{21}$

2025: 29 May Shift 2

The direction cosines of a line which makes equal angles with co-ordinate axes are:

2025: 19 May Shift 1

The shortest distance between lines $\frac{-x-3}{4} = \frac{y-6}{3} = \frac{z}{2}$ and $\frac{-x-2}{4} = \frac{y}{1} = \frac{z-7}{1}$ is:

2025: 14 May Shift 1

Consider the line $\vec{r} = \hat{i} - 2\hat{j} + 4\hat{k} + \lambda(-\hat{i} + 2\hat{j} - 4\hat{k})$

Match List-I with List-II

List-I	List-II
(A) A point on the given line	(I) $\left(\frac{-1}{\sqrt{21}}, \frac{2}{\sqrt{21}}, \frac{-4}{\sqrt{21}}\right)$
(B) direction ratios of the line	(II) $(4, -2, -2)$
(C) direction cosines of the line	(III) $(1, -2, 4)$
(D) direction ratios of a line perpendicular to given line	(IV) $(-1, 2, -4)$

Choose the correct answer from the options given below:

2024: 16 May Shift 1

The distance between the lines $\vec{r}=\hat{i}-2 \hat{j}+3 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$ and $\vec{r}=3 \hat{i}-2 \hat{j}+1 \hat{k}+\mu(4 \hat{i}+6 \hat{j}+12 \hat{k})$ is :

2026: 15 May Shift 1

The shortest distance between the lines $\dfrac{x+1}{7}=\dfrac{-y-1}{6}=\dfrac{z+1}{1}$ and $\dfrac{x-3}{1}=\dfrac{-y+5}{2}=\dfrac{z-7}{1}$ is

2025: 13 May Shift 2

If the lines $\frac{1-x}{3} = \frac{3y-6}{k} = \frac{3-z}{-2}$ and $\frac{1-x}{2k} = \frac{y-5}{3} = \frac{6-z}{5}$ are perpendicular to each other, then $k$ is equal to

Cosine of the acute angle between the lines x−32=y−21=z−52\frac{x-3}{2} = \frac{y-2}{1} = \frac{z-5}{2}2x−3​=1y−2​=2z−5​ and x−16=y−3−3=z+62\frac{x-1}{6} = \frac{y-3}{-3} = \frac{z+6}{2}6x−1​=−3y−3​=2z+6​ is

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CUET Mathematics Past Year Questions (Topic-Wise)

Cosine of the acute angle between the lines $\frac{x-3}{2} = \frac{y-2}{1} = \frac{z-5}{2}$ and $\frac{x-1}{6} = \frac{y-3}{-3} = \frac{z+6}{2}$ is