A vector perpendicular to a plane containing a triangle ABC having vertices as A(1,1,0), B(2,1,1) and C(0,3,2), is: Option 1: -2i - 3j + 2k Option 2: i + j + k Option 3: 2i - j - k Option 4: i - 2j + 2k

The answer is '-2i - 3j + 2k'. View full solution:

CUET Mathematics - A vector perpendicular to a plane containing a triangle ABC having vertices as A(1,1,0), B(2,1,1) and C(0,3,2), is: | PYQs + Solutions

CUET Mathematics 2022 23 Aug Shift 1

Algebra

Medium

$-2\hat{i} - 3\hat{j} + 2\hat{k}$

$\hat{i} + \hat{j} + \hat{k}$

$2\hat{i} - \hat{j} - \hat{k}$

$\hat{i} - 2\hat{j} + 2\hat{k}$

✅ Correct Option: 1

Related questions:

30 Aug Shift 1

If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ are two non zero vectors inclined at an angle $\theta$ , then identify the correct option out of the given options.

(a) $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot |\vec{b}|}$

(b) $\vec{a}$ and $\vec{b}$ are perpendicular, if $a_1 b_1 + a_2 b_2 + a_3 b_3 = 0$

(d) for $\theta = \pi$ , $\vec{a} \times \vec{b} = 0$

(e) $\cos\theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| \cdot |\vec{b}|}$

Choose the most appropriate answer from the options given below

29 May Shift 2

A vector of magnitude 9, which is perpendicular to both the vectors $(4\hat{i} - \hat{j} + 8\hat{k})$ and $(-\hat{j} + \hat{k})$ is

14 May Shift 2

Match List-I with List-II

Consider two vectors $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = -3\hat{i} - 6\hat{j} + 3\hat{k}$ , then

List-I	List-II
(A) Angle between $\vec{a}$ and $\vec{b}$ is	(I) $\cos^{-1}\left(\frac{1}{\sqrt{6}}\right)$
(B) Angle between $\vec{a}$ and $x$ -axis is	(II) $\cos^{-1}\left(\frac{2}{\sqrt{6}}\right)$
(C) Angle between $\vec{b}$ and $x$ -axis is	(III) $\pi$
(D) Angle between $\vec{a}$ and $y$ -axis is	(IV) $\cos^{-1}\left(-\frac{1}{\sqrt{6}}\right)$

Choose the correct answer from the options given below:

CUET Mathematics 2022 23 Aug Shift 1 PYQs

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