RD Sharma Class 12 Exercise 22.7 Algebra of vectors Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 22.7 Algebra of vectors Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 22.7 Algebra of vectors Solutions Maths - Download PDF Free Online

Lovekush kumar sainiUpdated on 24 Jan 2022, 06:20 PM IST

Class 12 RD Sharma chapter 22 exercise 22.7 solution is the best book which is available when it comes for the practice of mathematics for the class 12th CBSE students. Class 12th board exams are very important for every student. That's why RD Sharma class 12th exercise 22.7 has been recommended by great experts to grab each concept properly.

Also read - RD Sharma Solutions For Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter22 Algebra of vectors - Other Exercise

Algebra of Vectors Excercise: 22.7

Algebra of vectors exercise 22.7 question 1

Answer:
Point A, B and C are collinear
Hint:
Prove, that the position vectors are parallel to each other, so that they have one common point
Given:
Point A, B and C with position vectors $\vec{a}-2 \vec{b}+3 \vec{c}, 2 \vec{a}+3 \vec{b}-4 \vec{c} \: \text { and }-7 \vec{b}+10 \vec{c}$ respectively
Solution:
Let O be the point of origin for the position vectors
$\begin{aligned} &\therefore \overrightarrow{O A}=(\vec{a}-2 \vec{b}+3 \vec{c})-(0+0+0)=\vec{a}-2 \vec{b}+3 \vec{c} \\\\ &\text { Similarly } \overrightarrow{O B}=2 \vec{a}+3 \vec{b}-4 \vec{c} \text { and } \overrightarrow{O C}=-7 \vec{b}+10 \vec{c} \end{aligned}$
Now, $\overrightarrow{A B}=$ Position vector B – position vector A
$\begin{aligned} &=\overline{O B}-\overline{O A}\\ & \end{aligned}$
$=(2 \vec{a}+3 \vec{b}-4 \vec{c})-(\vec{a}-2 \vec{b}+3 \vec{c})\\$
$=[(2-1) \vec{a}]+[(3+2) \vec{b}]+[(-4-3) \vec{c}]\\$
$\overrightarrow{A B}=\vec{a}+5 \vec{b}-7 \vec{c}$ …(i)
$\overrightarrow{BC}$ Position of vector C – Position of vector B
$\begin{aligned} &\overrightarrow{B C}=[-7 \vec{b}+10 \vec{c}]-[2 \vec{a}+3 \vec{b}-4 \vec{c}] \\ & \end{aligned}$
$=[(0-2) \vec{a}]+[(-7-3) \vec{b}]+[(10+4) \vec{c}] \\$
$=-2 \vec{a}-10 \vec{b}+14 \vec{c}$
$=-2(\vec{a}+5 \vec{b}-7 \vec{c})$ …(ii)
Substituting (i) in (ii)
$\overrightarrow{B C}=-2(\overline{A B})$
Thus, $\overrightarrow{AB }\: and \: \overrightarrow{B C}$ are parallel vectors
As both vectors have one common point in B
$\overrightarrow{AB }\: and \: \overrightarrow{B C}$ are collinear
Thus, point A, B and C are collinear.

Algebra of vectors exercise 22.7 question 2(i)

Answer:
Point A, B and C are collinear
Hint:
Prove, that the position vectors are parallel to each other.
Given:
Point A, B and C with position vectors $\vec{a}, \vec{b} \text { and } 3 \vec{a}-2 \vec{b}$ respectively
Solution:
So, $\overrightarrow{A B}=$ Position of B vector – Position of A vector
$\overrightarrow{A B}=\vec{b}-\vec{a}$ …(i)
Similarly find $\overrightarrow{B C}$
$\overrightarrow{B C}=$ Position of C vector – Position of B vector
$\begin{aligned} &=(3 \vec{a}-2 \vec{b})-(\vec{b}) \\ & \end{aligned}$
$=3 \vec{a}-2 \vec{b}-\vec{b}$
$=3 \vec{a}-3 \vec{b}$
$\begin{aligned} &=3(\vec{a}-\vec{b}) \\ \end{aligned}$
$=-3(\vec{b}-\vec{a})$ …(ii)
Substituting (i) in (ii)
Thus, $\overrightarrow{B C}=-3 \overrightarrow{A B}$
So we say that $\overrightarrow{B C}\; \text{and}\; \overrightarrow{A B}$ parallel to each other.
Here B is a common point in both vector
Thus, point A, B and C are collinear.

Algebra of vectors exercise 22.7 question 2(ii)

Answer:
Point A, B and C are collinear
Hint:
Prove, that the position vectors are parallel to each other
Given:
$\vec{a}, \vec{b} \& \vec{c}$ Are non-coplanar vectors
Solution:
Let, points A, B and C be the position vectors for $\vec{a}+\vec{b}+\vec{c}, 4 \vec{a}+3 \vec{b}, 10 \vec{a}+7 \vec{b}-2 \vec{c}$ respectively
$\overrightarrow{A B}$ Position vector B – Position vector A
$=(4 \vec{a}+3 \vec{b})-(\vec{a}+\vec{b}+\vec{c}) \\$
$=3 \vec{a}+2 \vec{b}+(-\vec{c}) \\$
$\begin{aligned} & &\overrightarrow{A B}=3 \vec{a}+2 \vec{b}-\vec{c} \end{aligned}$ …(i)
Similarly,
$\begin{aligned} &\overrightarrow{B C}=(10 \vec{a}+7 \vec{b}-2 \vec{c})-(4 \vec{a}+3 \vec{b})\\ & \end{aligned}$
$=6 \vec{a}+4 \vec{b}-2 \vec{c}\\$
$=2(3 \vec{a}+2 \vec{b}-\vec{c})$ …(ii)
Substitute (i) in (ii)
$\therefore \overline{B C}=2 \overrightarrow{A B}$
Hence, $\overrightarrow{A B} \& \overline{B C}$ are parallel vectors.
As B point is common in both vectors
Thus, A, B and C are collinear points.

Algera of vectors excercise 22.7 question 3

Answer:
Point A, B and C are collinear
Hint:
Obtain the parallel vectors and if any common points is there they are parallel to each other
Given:
Three vectors are given.
Solution:
Let,
$\begin{aligned} &A=\hat{i}+2 \hat{j}+3 \hat{k}\\ & \end{aligned}$
$B=3 \hat{i}+4 \hat{j}+7 \hat{k}\\$
$C=-3 \hat{i}-2 \hat{j}-5 \hat{k}\\$
$\therefore \overrightarrow{A B}=(3 \hat{i}+4 \hat{j}+7 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})\\$
$=(3-1) \hat{i}+(4-2) \hat{j}+(7-3) \hat{k}\\$
$=2 \hat{i}+2 \hat{j}+4 \hat{k}$ … (i)
Similarly
$\begin{aligned} &\overrightarrow{B C}=(-3 \hat{i}-2 \hat{j}-5 \hat{k})-(3 \hat{i}+4 \hat{j}+7 \hat{k})\\ & \end{aligned}$
$=(-3-3) \hat{i}+(-2-4) \hat{j}+(-5-7) \hat{k}\\$
$=-6 \hat{i}-6 \hat{j}-12 \hat{k}\\$
$=-6(\hat{i}+\hat{j}+2 \hat{k})\\$
$\begin{aligned} &\overrightarrow{B C}=-3(2 \hat{i}+2 \hat{j}+4 \hat{k}) \end{aligned}$ … (ii)
Put (i) in (ii)
Thus, $\overrightarrow{B C}=-3 \overrightarrow{A B}$
From above result, B is the common point in $\overline{A B} \& \overline{B C}$
A, B and C points are collinear.

Algera of vectors excercise 22.7 question 4

Answer:
$8$
Hint:
As the given vectors are collinear, then use formula $\overrightarrow{A B}=\lambda \overrightarrow{B C}$
Given:
Three vectors are collinear.
Solution:
Let,
$\begin{aligned} &A=10 \hat{i}+3 \hat{j} \\ \end{aligned}$
$B=12 \hat{i}-5 \hat{j} \\$
$C=a \hat{i}+11 \hat{j}$
So the above three points A, B and C represents three vectors collinear to each other
$\begin{aligned} &\overrightarrow{A B}=(12 \hat{i}-5 \hat{j})-(10 \hat{i}+3 \hat{j}) \\ & \end{aligned}$
$=(12-10) \hat{i}+(-5-3) \hat{j} \\$
$=2 \hat{i}-8 \hat{j} \\$
$\overrightarrow{B C}=(a \hat{i}+11 \hat{j})-(12 \hat{i}-5 \hat{j}) \\$
$=(a-12) \hat{i}+(11+5) \hat{j} \\$
$=(a-12) \hat{i}+16 \hat{j}$
As, A,B and C are collinear, then there exists $\overrightarrow{A B}=\lambda \overrightarrow{B C}$
$\therefore(2 \hat{i}-8 \hat{j})=\lambda(a-12) \hat{i}+\lambda 16 \hat{j}$
Comparing
$\begin{aligned} &-8 j=\lambda 16 \hat{j} \\ & \end{aligned}$
$-8=\lambda 16 \\$
$\lambda=\frac{-1}{2} \\$
$2 \hat{i}=\lambda(a-12) \hat{i} \\$
$2=\lambda(a-12)$
Put the value of $\lambda$
$\begin{aligned} &2=\frac{-1}{2}(a-12) \\ \end{aligned}$
$-4=a-12 \\$
$a=8$

Algera of vectors excercise 22.7 question 5

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.

Algebra of vectors exercise 22.7 question 6

Answer:
A, B and C are collinear points
Hint:
From the origin, point O towards a point ‘A’. $\overrightarrow{O A}=-\overrightarrow{A O}$

Given:
A, B and C are collinear
Solution:
As per shown figure above
$\begin{aligned} \overrightarrow{O A} &=-\overrightarrow{A O} \\ \overrightarrow{O B} &=-\overrightarrow{B O} \\ \overrightarrow{O C} &=-\overrightarrow{C O} \end{aligned}$
Where O must be the origin
We have,
$\begin{aligned} &\overrightarrow{A O}+\overrightarrow{O B}=\overrightarrow{B O}+\overrightarrow{O C} \\ &\overrightarrow{A O}-\overrightarrow{B O}=\overrightarrow{O C}-\overrightarrow{O B} \\ &(-\overrightarrow{O A})-(-\overrightarrow{O B})=\overrightarrow{O C}-\overrightarrow{O B} \\ &\overrightarrow{O B}-\overrightarrow{O A}=\overrightarrow{O C}-\overrightarrow{O B} \end{aligned}$
From above figure we can say
$\overrightarrow{A B}=\overrightarrow{B C}$

Algebra of vectors exercise 22 .7 question 7

Answer:
Given position vectors are collinear
Hint:
Just prove that one vector can be represented in form of another
Given:
$2 \hat{i}-3 \hat{j}+4 \hat{k} \text { And }-4 \hat{i}+6 \hat{j}-8 \hat{k}$
Solution:
Let,
$\begin{aligned} &\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k} \\ & \end{aligned}$ … (i)
$\vec{b}=-4 \hat{i}+6 \hat{j}-8 \hat{k} \\$
$\vec{b}=-2(2 \hat{i}-3 \hat{j}+4 \hat{k}) \\$
$\vec{b}=-2 \vec{a}$ (From (i))
Thus $\vec{a} \& \vec{b}$ are collinear.

Algebra of vectors exercise 22.7 Question 8

Answer:
$1$
Hint:
Represent point in form of vectors and use $\overrightarrow{A B}=\lambda \overrightarrow{B C}$
Given:
Points $A(m,-1), B(2,1), C(4,5)$ are collinear
Solution:
$\begin{aligned} &A=m \hat{i}-\hat{j} \\ & \end{aligned}$
$B=2 \hat{i}+\hat{j} \\$
$C=4 \hat{i}+5 \hat{j} \\$
$\overrightarrow{A B}=(2 \hat{i}+\hat{j})-(m \hat{i}-\hat{j}) \\$
$=(2-m) \hat{i}+(1+1) \hat{j} \\$
$=(2-m) \hat{i}+2 \hat{j} \\$
$\overrightarrow{B C}=(4 \hat{i}+5 \hat{j})-(2 \hat{i}+\hat{j}) \\$
$=(4-2) \hat{i}+(5-1) \hat{j} \\$
$=2 \hat{i}+4 \hat{j}$
If they are collinear then
$\begin{aligned} &\overrightarrow{A B}=\lambda \overrightarrow{B C} \\ &(2-m) \hat{i}+2 \hat{j}=\lambda(2 \hat{i}+4 \hat{j}) \end{aligned}$
Comparing
$\begin{aligned} &2=\lambda(4) \\ &\lambda=\frac{1}{2} \\ &(2-m)=\lambda(2) \end{aligned}$
Put the value of $\lambda$
$\begin{aligned} &2-m=\frac{1}{2} \times 2 \\ &2-m=1 \\ &m=2-1 \\ &m=1 \end{aligned}$

Algebra of vectors exercise 22.7 question 9

Answer:
Given points are collinear
Hint:
If the points are collinear then they are parallel.
Given:
$(3,4),(-5,16) \&(5,1)$ Are collinear
Solution:
Let,
$\begin{aligned} &A=3 \hat{i}+4 \hat{j} \\ &B=-5 \hat{i}+16 \hat{j} \\ &C=5 \hat{i}+\hat{j} \end{aligned}$
$\therefore \overrightarrow{A{B}}=$ Position of B – Position of A
$=(-5 \hat{i}+16 \hat{j})-(3 \hat{i}+4 \hat{j}) \\$
$=(-5-3) \hat{i}+(16-4) \hat{j} \\$
$=-8 \hat{i}+12 \hat{j} \\$
$\begin{aligned} & &=-4(2 \hat{i}-3 \hat{j}) \end{aligned}$ … (i)
$\overrightarrow{B C}$ Position of C – Position of B
$=(\hat{5} \hat{i}+\hat{j})-(-5 \hat{i}+16 \hat{j}) \\$
$=(5+5) \hat{i}+(1-16) \hat{j} \\$
$=10 \hat{i}-15 \hat{j} \\$
$\begin{aligned} & &=5(2 \hat{i}-3 \hat{j}) \end{aligned}$ … (ii)
From (i) and (ii)
$\overrightarrow{B C}=\frac{-5}{4} \overrightarrow{A B}$
Thus $\overrightarrow{B C} \& \overrightarrow{A B}$ are parallel to each and B being the common point
A, B and C are collinear
Thus, $(3,4),(-5,16) \&(5,1)$ are collinear.

Algebra of vectors exercise 22.7 question 10

Answer:
$9$
Hint:
If one vector is scalar product of another vector, then they are parallel and collinear.
Given:
$\begin{aligned} &\vec{a}=2 \hat{i}-3 \hat{j} \\ &\vec{b}=-6 \hat{i}+m \hat{j} \end{aligned}$ Are collinear
Solution:
$\begin{aligned} &\vec{a}=2 \hat{i}-3 \hat{j} \\ & \end{aligned}$
$\vec{b}=-6 \hat{i}+m \hat{j} \\$
$\vec{b}=\lambda \vec{a} \\$
$(-6 \hat{i}+m \hat{j})=\lambda(2 \hat{i}-3 \hat{j})$
Comparing we get
$-6=2 \lambda \\$
$\lambda=-3 \\$ … (i)
$m=-3 \lambda \\$
$m=-3(-3) \\$ … (From (i))
$\begin{aligned} & &m=9 \end{aligned}$
$\therefore$ Value of m is 9

Algebra of vectors exercise 22.7 question 11

Answer:
$2:3$
Hint:
When, in a three collinear points A, B and C, if B divides AC then formula is $\frac{n \vec{a}+m \vec{b}}{n+m}$
Given:
$A(1,-2,-8), B(5,0,-2) \& C(11,3,7)$ Are collinear points
Solution:
$\begin{aligned} &A=\hat{i}-2 \hat{j}-8 \hat{k} \\ &B=5 \hat{i}-2 \hat{k} \\ &C=11 \hat{i}+3 \hat{j}+7 \hat{k} \\ &\overrightarrow{A B}=(5 \hat{i}-2 \hat{k})-(\hat{i}-2 \hat{j}-8 \hat{k}) \\ &=(5-1) \hat{i}+(0+2) \hat{j}+(-2+8) \hat{k} \end{aligned}$
$=4 \hat{i}+2 \hat{j}+6 \hat{k} \\$ … (i)
$\overrightarrow{B C}=(11 \hat{i}+3 \hat{j}+7 \hat{k})-(5 \hat{i}-2 \hat{k}) \\$
$=(11-5) \hat{i}+(3-0) \hat{j}+(7+2) \hat{k} \\$
$\begin{aligned} & &=6 \hat{i}+3 \hat{j}+9 \hat{k} \end{aligned}$ … (ii)
Rewriting (i) and (ii)
$\overrightarrow{A B}=2(2 \hat{i}+\hat{j}+3 \hat{k}) \\$ … (iii)
$\begin{aligned} & &\overrightarrow{B C}=3(2 \hat{i}+\hat{j}+3 \hat{k}) \end{aligned}$ … (iv)
From (iii)
$\overrightarrow{B C}=\frac{3}{2} \overline{A B}$
Thus, $\overrightarrow{B C} \& \overrightarrow{A B}$ are parallel to each other
Therefore they are collinear
Now,

Now, if B is dividing AC internally into m and n then,
$\begin{aligned} &B=\left(\frac{11 m+n}{m+n}\right) \hat{i}+\left(\frac{3 m-2 n}{m+n}\right) \hat{j}+\left(\frac{7 m-8 n}{m+n}\right) \hat{k} \\\\ &5 \hat{i}-2 \hat{k}=\left(\frac{11 m+n}{m+n}\right) \hat{i}+\left(\frac{3 m-2 n}{m+n}\right) \hat{j}+\left(\frac{7 m-8 n}{m+n}\right) \hat{k} \end{aligned}$
Comparing,
$\begin{aligned} &5=\frac{11 m+n}{m+n} \\ &5 m+5 n=11 m+n \\ &\therefore 6 m-4 n=0 \end{aligned}$ … (v)
$\begin{gathered} 0=\frac{3 m-2 n}{m+n} \\ 0=3 m-2 n \\ 3 m-2 n=0 \end{gathered}$ … (vi)
$\begin{aligned} &-2=\frac{7 m-8 n}{m+n} \\ &-2 m-2 n=7 m-8 n \\ &9 m-6 n=0 \end{aligned}$ … (vii)
Now, solving equation (v) and (VI) by elimination method
$\begin{aligned} &6 m-4 n=0 \\ &2(3 m-2 n)=0 \\ &3 m-2 n=0 \\ &3 m=2 n \end{aligned}$
If we suppose n=1 then
$3 m=2(1) \\$
$\begin{aligned} & &m=\frac{2}{3} \end{aligned}$ For all the three equation.

Algebra of vectors exercise 22.7 question 12

Answer:
AB and CD intersect at the point P
Hint:
If $\overrightarrow{A B} \& \overrightarrow{C D}$ intersect at P, it makes A-P-B collinear and C-P-D collinear. So prove they are collinear
Given:
$\begin{aligned} &A=-2 \hat{i}+3 \hat{j}+5 \hat{k} \\ & \end{aligned}$
$B=7 \hat{i}+0 \hat{j}-\hat{k}=7 \hat{i}-\hat{k} \\$
$C=-3 \hat{i}-2 \hat{j}-5 \hat{k} \\$
$D=3 \hat{i}+4 \hat{j}+7 \hat{k} \\$
$P=1 \hat{i}+2 \hat{j}+3 \hat{k}$
Solution:
$\begin{aligned} &A=-2 \hat{i}+3 \hat{j}+5 \hat{k} \\ &B=7 \hat{i}+0 \hat{j}-\hat{k}=7 \hat{i}-\hat{k} \\ &C=-3 \hat{i}-2 \hat{j}-5 \hat{k} \\ &D=3 \hat{i}+4 \hat{j}+7 \hat{k} \\ &P=1 \hat{i}+2 \hat{j}+3 \hat{k} \end{aligned}$
$\overrightarrow{A P}=$ Position vector of P – Position vector of A
$\begin{aligned} &=(\hat{i}+2 \hat{j}+3 \hat{k})-(-2 \hat{i}+3 \hat{j}+5 \hat{k})\\ &=(1+2) \hat{i}+(2-3) \hat{j}+(3-5) \hat{k}\\ &=3 \hat{i}-\hat{j}-2 \hat{k} \end{aligned}$ … (i)
$\overrightarrow{P B}$ Position vector of B – Position vector of P
$\begin{aligned} &=(7 \hat{i}-\hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})\\ & \end{aligned}$
$=(7-1) \hat{i}+(0-2) \hat{j}+(-1-3) \hat{k}\\$
$=6 \hat{i}-2 \hat{j}-4 \hat{k}\\$
$=2(3 \hat{i}-\hat{j}-2 \hat{k}) \\$ cgfy… (ii)
$\overline{P B}=2 \overrightarrow{A P}$
Thus, A-P-B are collinear
$\overrightarrow{C P}=$ Position vector of P – Position vector C
$\begin{aligned} &=(\hat{i}+2 \hat{j}+3 \hat{k})-(-3 \hat{i}-2 \hat{j}-5 \hat{k}) \\ &=(1+3) \hat{i}+(2+2) \hat{j}+(3+5) \hat{k} \\ &=4 \hat{i}+4 \hat{j}+8 \hat{k} \\ &=4(\hat{i}+\hat{j}+2 \hat{k}) \end{aligned}$
$\overrightarrow{P D}=$ Position vector of D – Position vector of P
$\begin{aligned} &=(3 \hat{i}+4 \hat{j}+7 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k}) \\ &=(3-1) \hat{i}+(4-2) \hat{j}+(7-3) \hat{k} \\ &=2 \hat{i}+2 \hat{j}+4 \hat{k} \\ &=2(\hat{i}+\hat{j}+2 \hat{k}) \\ &\overrightarrow{P D}=\frac{2}{4} \overrightarrow{C P} \\ &\overrightarrow{P D}=\frac{1}{2} \overrightarrow{C P} \end{aligned}$
Thus, $\overrightarrow{P D} \& \overline{C P}$ are collinear
If A-P-B are collinear and also C-P-D are collinear
Hence $\overrightarrow{A B} \text { and } \overrightarrow{C D}$ intersect at P

Algebra of vectors exercise 22.7 Question 13

Answer:
$-2$
Hint:
Use the formula $\overrightarrow{A B}=K \overrightarrow{B C}$ to prove it is collinear
Given:
$\begin{aligned} &A=\lambda \hat{i}-10 \hat{j}+3 \hat{k} \\ &B=\hat{i}-\hat{j}+3 \hat{k} \\ &C=3 \hat{i}+5 \hat{j}+3 \hat{k} \end{aligned}$
Solution:
$\begin{aligned} &A=\lambda \hat{i}-10 \hat{j}+3 \hat{k} \\ &B=\hat{i}-\hat{j}+3 \hat{k} \\ &C=3 \hat{i}+5 \hat{j}+3 \hat{k} \end{aligned}$
$\begin{aligned} &\overrightarrow{A B}=(\hat{i}-\hat{j}+3 \hat{k})-(\lambda \hat{i}-10 \hat{j}+3 \hat{k}) \\ &=(1-\lambda) \hat{i}+(-1+10) \hat{j}+(3-3) \hat{k} \\ &=(1-\lambda) \hat{i}+9 \hat{j} \end{aligned}$
$\begin{aligned} &\overrightarrow{B C}=(3 \hat{i}+5 \hat{j}+3 \hat{k})-(\hat{i}-\hat{j}+3 \hat{k}) \\ &=(3-1) \hat{i}+(5+1) \hat{j}+(3-3) \hat{k} \\ &=2 \hat{i}+6 \hat{j} \\ &\overline{A B}=K \overline{B C} \\ &(1-\lambda) \hat{i}+9 \hat{j}=K(2 \hat{i}+6 \hat{j}) \end{aligned}$
Comparing

$\begin{aligned} &6 K=9 \\ &\therefore K=\frac{9}{6}=\frac{3}{2} \\ &K=\frac{3}{2} \\ &1-\lambda=2 K \end{aligned}$
$\begin{aligned} &1-\lambda=2\left(\frac{3}{2}\right) \\ &1-\lambda=3 \\ &1-3=\lambda \\ &\lambda=-2 \end{aligned}$

RD Sharma Class 12th Exercise 22.7 material is a good alternative for exam preparation as it follows the CBSE syllabus and covers all topics. The RD Sharma Class 12 solution of Algebra of Vectors exercise 22.7 contains 14 questions that help students get a good idea about the concept used.

Dot and cross product of two vectors
Unit vectors and position vector
Collinear vectors
Non-collinear vectors
Non-planar vectors

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