SATHEE: 3D Geometry 3 Question 13

3D Geometry 3 Question 13

####13. The magnitude of the projection of the vector $2 \hat{i} + 3 \hat{j} + \hat{k}$ on the vector perpendicular to the plane containing the vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 2 \hat{j} + 3 \hat{k}$ , is

(2019 Main, 8 April I)

(a) $\frac{\sqrt{3}}{2}$

(b) $\sqrt{6}$

(d) $\sqrt{\frac{3}{2}}$

Show Answer

Answer:

(d)

Solution:

The normal vector to the plane containing the vectors $(\hat{i} + \hat{j} + \hat{k})$ and $(\hat{i} + 2 \hat{j} + 3 \hat{k})$ is $n = (\hat{i} + \hat{j} + \hat{k}) \times (\hat{i} + 2 \hat{j} + 3 \hat{k})$

$\begin{aligned} \begin{array}{lll} \hat{i} \hat{j} \hat{k} \end{array} \\ = & 1 1 1 \\ 1 2 3 \\ = \hat{i} (3 - 2) - \hat{j} (3 - 1) + \hat{k} (2 - 1) = \hat{i} - 2 \hat{j} + \hat{k} \end{aligned}$

Now, magnitude of the projection of vector $2 \hat{i} + 3 \hat{j} + \hat{k}$ on normal vector $n$ is

$\begin{aligned} \frac{| (2 \hat{i} + 3 \hat{j} + \hat{k}) \cdot n |}{| n |} & = \frac{| (2 \hat{i} + 3 \hat{j} + \hat{k}) \cdot (\hat{i} - 2 \hat{j} + \hat{k}) |}{\sqrt{1 + 4 + 1}} \\ = \frac{| 2 - 6 + 1 |}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \sqrt{\frac{3}{2}} units \end{aligned}$

3D Geometry 3 Question 13

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

3D Geometry 3 Question 13

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu