SATHEE: Vectors 2 Question 18

Vectors 2 Question 18

18. The unit vector perpendicular to the plane determined by $P (1, - 1, 2), Q (2, 0, - 1)$ and $R (0, 2, 1)$ is . .

(1983, 2M)

Show Answer

Answer:

Correct Answer: 18. (b, d)

Solution:

A unit vector perpendicular to the plane determined by

$\begin{aligned} P, Q, R & = \pm \frac{(\vec{P Q}) \times (\vec{P R})}{| \vec{P Q} \times \vec{P R} |} \\ ∴ Unit vector & = \pm \frac{(\vec{P Q}) \times (\vec{P R})}{| \vec{P Q} \times \vec{P R} |} \end{aligned}$

where, $\vec{P Q} = \hat{i} + \hat{j} - 3 \hat{k}$

and $\vec{P R} = - \hat{i} + 3 \hat{j} - \hat{k}$

$∴ \vec{P Q} \times \vec{P R} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} 1 & 1 & - 3 - 1 & 3 & - 1 \end{array} |$

$= \hat{i} (- 1 + 9) - \hat{j} (- 1 - 3) + \hat{k} (3 + 1)$

$= 8 i + 4 j + 4 \hat{k}$

$\Rightarrow | \vec{P Q} \times \vec{P R} | = 4 \sqrt{4 + 1 + 1} = 4 \sqrt{6}$

$∴ \frac{\vec{P Q} \times \vec{P R}}{| \vec{P Q} \times \vec{P R} |} = \pm \frac{4 (2 \hat{i} + \hat{j} + \hat{k})}{4 \sqrt{6}}$

$= \pm \frac{(2 \hat{i} + \hat{j} + \hat{k})}{\sqrt{6}}$

Vectors 2 Question 18

18. The unit vector perpendicular to the plane determined by $P (1, - 1, 2), Q (2, 0, - 1)$ and $R (0, 2, 1)$ is . .

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Vectors 2 Question 18

18. The unit vector perpendicular to the plane determined by P(1,−1,2),Q(2,0,−1) and R(0,2,1) is . .

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

18. The unit vector perpendicular to the plane determined by $P (1, - 1, 2), Q (2, 0, - 1)$ and $R (0, 2, 1)$ is . .