The direction cosines of the line which is perpendicular to the lines with direction ratios 1,-2,-2 and 0,2,1 are : Option 1: 2/3,-1/3, 2/3 Option 2: -2/3,-1/3, 2/3 Option 3: 2/3,-1/3,-2/3 Option 4: 2/3, 1/3, 2/3

Answer: 2/3,-1/3, 2/3. Solution: The two given lines have direction ratios: - Line 1: (1, -2, -2) - Line 2: (0, 2, 1) When two vectors are given, their cross product gives a vector perpendicular to both. Let's calculate the cross product (1, -2, -2) x (0, 2, 1) Using the formula a x b = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1): = ((-2)(1) - (-2)(2), (-2)(0) - (1)(1), (1)(2) - (-2)(0)) = (-2 - (-4), 0 - 1, 2 - 0) = (2, -1, 2) Direction cosines are normalized direction ratios. To normalize, divide each component by the magnitude. Magnitude: \, (2, -1, 2)\, = √(2^2 + (-1)^2 + 2^2) = √9 = 3 Therefore, the direction cosines are: (2/3, -1/3, 2/3)

Geometry

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The direction cosines of the line which is perpendicular to the lines with direction ratios $1,-2,-2$ and $0,2,1$ are :

$\frac{2}{3},-\frac{1}{3}, \frac{2}{3}$

$-\frac{2}{3},-\frac{1}{3}, \frac{2}{3}$

$\frac{2}{3},-\frac{1}{3},-\frac{2}{3}$

$\frac{2}{3}, \frac{1}{3}, \frac{2}{3}$

✅ Correct Option: 1

The two given lines have direction ratios:

Line 1: $(1, -2, -2)$ $(1, - 2, - 2)$
- Line 2: $(0, 2, 1)$

When two vectors are given, their cross product gives a vector perpendicular to both.

Let's calculate the cross product $(1, -2, -2) \times (0, 2, 1)$

Using the formula $\vec{a} \times \vec{b} = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1)$ :

$= ((-2)(1) - (-2)(2), (-2)(0) - (1)(1), (1)(2) - (-2)(0))$

$= (-2 - (-4), 0 - 1, 2 - 0)$

$= (2, -1, 2)$

Direction cosines are normalized direction ratios. To normalize, divide each component by the magnitude.

Magnitude: $\|(2, -1, 2)\| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3$

Therefore, the direction cosines are:

$\left(\dfrac{2}{3}, -\dfrac{1}{3}, \dfrac{2}{3}\right)$

2022: 17 Aug Shift 2

The equation of the plane, parallel to the plane $3x + 4y - 12z = 3$ and passes through $(1, 1, -1)$ , 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:

2023: 15 June Shift 2

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

2025: 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

2025: 15 May Shift 1

If $\alpha, \beta$ and $\gamma$ are angle of inclinations of a line with x, y and z axes respectively, then the value of $2(\cos 2\alpha + \cos 2\beta + \cos 2\gamma)$ is

The direction cosines of the line which is perpendicular to the lines with direction ratios 1,−2,−21,-2,-21,−2,−2 and 0,2,10,2,10,2,1 are :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The direction cosines of the line which is perpendicular to the lines with direction ratios $1,-2,-2$ and $0,2,1$ are :