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Need solution for RD Sharma Maths Class 12 Chapter 22 Algera of Vectors Excercise 22.7 Question 5

Answers (1)

Answer:

Given position vectors are collinear for all the real values of $\lambda$

Hint:

To be collinear, vectors must be a multiple of another.

Given:

$\vec{a}, \vec{b}$ Are non collinear vectors .

Solution:

Let, position vectors of points X, Y and Z are $\vec{a}+\vec{b}, \vec{a}-\vec{b} \& \vec{a}+\lambda \vec{b}$ respectively.

Then, $\overrightarrow{XY}$ Position of vector Y – Position of vector X

$\begin{aligned} &=(\vec{a}-\vec{b})-(\vec{a}+\vec{b}) \\ & \end{aligned}$

$=-2 \vec{b}$

Similarly

$\overrightarrow{YZ}$ Position of vector Z – Position of vector Y

$\begin{aligned} &=(\vec{a}+\lambda \vec{b})-(\vec{a}-\vec{b})\\ \end{aligned}$

$=\lambda \vec{b}+\vec{b}\\$

$=\vec{b}(\lambda+1)$ … (i)

Now,

$\overrightarrow{Z X}$ Position of vector X – Position of Z

$\begin{aligned} &=(\vec{a}+\vec{b})-(\vec{a}+\lambda \vec{b}) \\ \end{aligned}$

$=\vec{b}-\lambda \vec{b} \\$

$=\vec{b}(1-\lambda) \\$

$=-\vec{b}(\lambda-1)$ …(ii)

From (i) and (ii)

$\overrightarrow{Z X}=-K \overrightarrow{Y Z}$

Here point Z is common

Thus, we can say that for all real values of $\lambda$ given position vectors are parallel

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