Let \veca=\hati-\hatj and \vecb=\hati+\hatj+\hatk and \vecc be a vector such ...

Aspire's Library

A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations - NIMCET

Best NIMCET Coaching
Aspire Study
Online Classes, Classroom Classes
and More.

Aspire Study Library

Question Id : 11296 | Context :CUET 2022

Question

Let

$\vec{a}=\hat{i}-\hat{j}$ and

$\vec{b}=\hat{i}+\hat{j}+\hat{k}$ and

$\vec{c}$ be a vector such that

$(\vec{a} \times \vec{c})+\vec{b}=0$ and

$\vec{a}.\vec{c}=4$ , then

$|\vec{c}|^2$ is equal to

🎥 Video solution / Text Solution of this question is given below:

Solution:

Given vectors:

$\vec{a} = \hat{i} - \hat{j}$
$\vec{b} = \hat{i} + \hat{j} + \hat{k}$
And $(\vec{a} \times \vec{c}) + \vec{b} = 0$
$\vec{a} \cdot \vec{c} = 4$

From

$(\vec{a} \times \vec{c}) + \vec{b} = 0$ , we get:

$\vec{a} \times \vec{c} = -\vec{b}$ Let

$\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$ . The cross product

$\vec{a} \times \vec{c}$ is:

$\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ x & y & z \end{vmatrix}$ Expanding this determinant:

$\vec{a} \times \vec{c} = (z \hat{i} + z \hat{j} + (x + y) \hat{k})$ Setting

$\vec{a} \times \vec{c} = -\vec{b}$ , we get:

$z = -1, \quad z = -1, \quad x + y = -1$ Therefore:

$x + y = -1$ Now, from

$\vec{a} \cdot \vec{c} = 4$ :

$\vec{a} \cdot \vec{c} = 1 \cdot x + (-1) \cdot y = 4$ Simplifying:

$x - y = 4$ Solving the system of equations:

$x + y = -1$

$x - y = 4$ Adding the two equations:

$2x = 3 \quad \Rightarrow \quad x = \frac{3}{2}$ Substituting into

$x + y = -1$ :

$\frac{3}{2} + y = -1 \quad \Rightarrow \quad y = -\frac{5}{2}$ Now,

$\vec{c} = \frac{3}{2} \hat{i} - \frac{5}{2} \hat{j} - \hat{k}$ . To find

$|\vec{c}|^2$ , we compute:

$|\vec{c}|^2 = \left( \frac{3}{2} \right)^2 + \left( -\frac{5}{2} \right)^2 + (-1)^2 = \frac{9}{4} + \frac{25}{4} + 1$

$|\vec{c}|^2 = \frac{9 + 25 + 4}{4} = \frac{38}{4} = 9.5$
Final Answer:

$\boxed{9.5}$

📲 Install Aspire Study App

Get MCA exam-wise questions, solutions, test series & updates instantly on your phone.

🚀 Install Now

Aspire Study Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Aspire Study Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Ask Your Question or Put Your Review.

Feedback

Info.

Information

Aspire's Library

Aspire Study Library

📲 Share This Question

Question

Solution:

📲 Install Aspire Study App

Info.

Confirm Delete

Edit