SATHEE: 3D Geometry 3 Question 31

3D Geometry 3 Question 31

####31. Let $u$ be a vector coplanar with the vectors $a = 2 \hat{i} + 3 \hat{j} - \hat{k}$ and $b = \hat{j} + \hat{k}$ . If $u$ is perpendicular to $a$ and $u \cdot b = 24$ , then $| u |^{2}$ is equal to

(2018 Main)

(a) 336

(b) 315

(d) 84

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Answer:

Correct Answer: 31. (a)

Solution:

Key Idea If any vector $x$ is coplanar with the vector $y$ and $z$ , then $x = λ y + μ z$

Here, $u$ is coplanar with $a$ and $b$ .

$∴ u = λ a + μ b$

Dot product with a, we get

$u \cdot a = λ (a \cdot a) + μ (b \cdot a) \Rightarrow 0 = 14 λ + 2 μ$

$[∵ a = 2 \hat{i} + 3 \hat{j} - \hat{k}, b = \hat{j} + \hat{k}, u \cdot a = 0]$

Dot product with $b$ , we get

$\begin{aligned} u \cdot b = λ (a \cdot b) + μ (b \cdot b) \\ 24 = 2 λ + 2 μ \end{aligned}$

Solving Eqs. (i) and (ii), we get

$λ = - 2, μ = 14$

Dot product with $u$ , we get

$\begin{aligned} | u |^{2} = λ (u \cdot a) + μ (u \cdot b) \\ | u |^{2} = - 2 (0) + 14 (24) \Rightarrow | u |^{2} = 336 \end{aligned}$

3D Geometry 3 Question 31

Answer:

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3D Geometry 3 Question 31

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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