JEE Main Class 11 Syllabus 2025 PDF for Paper 1 and 2

Linear Combination of Vectors - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

15 Questions around this concept.

Solve by difficulty

If $\vec{a},\vec{b},\vec{c}$ are non-coplanar vectors and $\lambda$ is a real number, then the vectors $\vec{a}+2\vec{b}+3\vec{c},\; \lambda\vec{b}+4\vec{c}\; \; and\; \; (2\lambda -1)\vec{c}$ are non-coplanar for

all except two values of $\lambda$

all except one values of $\lambda$

all values of $\lambda$

no value of $\lambda$

View Solution

Concepts Covered - 2

Linear Combination of Vectors

$\\ {\text {A vector } \vec{r} \text { is said to be a linear combination of vectors } \vec{a}_1, \;\vec{a}_2},\;\vec{a}_3 \dots\ldots \;, \vec{a}_n \\\text {if there exist scalars } \lambda_1,\;\lambda_2,\;\lambda_3,\;\ldots\ldots,\lambda_n \text { such that }\\\\\vec {r}=\lambda_1\;\vec{a}_1+\lambda_2\;\vec{a}_2+\lambda_3\;\vec{a}_3+\ldots\ldots+\lambda_n\;\vec{a}_n$

For example:

$\\\mathrm{vectors\;\;}\vec r_1=\vec a+2\vec b+3\vec c\;\;\text{and }\;\;\vec r_2=\vec a-4\vec b-9\vec c\;\;\text{are linear combination of}\\\text{the vectors }\vec a,\;\vec b\;\;and\;\;\vec c.$

Linear Independent Vectors

$\\\mathrm{A\;system\;of\;vectors\;\;} \vec{a}_1, \;\vec{a}_2,\;\vec{a}_3 \dots\ldots \;, \vec{a}_n\text{ is said to be linearly independent,\;if}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\lambda_1\;\vec{a}_1+\lambda_2\;\vec{a}_2+\lambda_3\;\vec{a}_3+\ldots\ldots+\lambda_n\;\vec{a}_n=0\\\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lambda_1\;=\lambda_2\;=\lambda_3\;=\ldots\ldots=\lambda_n\;=0$

It can be easily verified that

1. A pair of non-collinear vectors ( say a₁ and a₂ ) are linearly independent.

$\\\mathrm{Let \;}\vec a_1\;\text{and }\vec a_2\;\text{are two non-collinear vectors such that }\;\lambda_1\vec a_1+\lambda_2\vec a_2=0.\\\mathrm{Let,\;\;\;\lambda_1,\;\lambda_2\;\neq 0}\\\mathrm{\Rightarrow \;\;\;\;\;\;\;}\vec a_1=-\frac{\lambda_2}{\lambda_1}\;\vec a_2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(i)\\\text{Now, }-\frac{\lambda_2}{\lambda_1}\text{ is a scalar, because }\lambda_1\;\text{and }\lambda_2\;\text{ are scalars.}\\\mathrm{Hence,\;eq\;(i)\; }\text { expresses } \vec a_1 \text{ as product of } \vec a_2 \text{ by a scalar, so that }\vec a_1 \text{ and }\vec a_2 \text{ are collinear.}\\\mathrm{Which\;contradict\; the \;given \;fact\;because\;\overrightarrow{a_1}\;and\;\overrightarrow{a_2}\;are \;given\;to\;be\;non-collinear.}\\\mathrm{Th\;thus,\;\;our\;supposition\;that\;\lambda_1\neq0\;and\;\lambda_2\neq0\;\;is\;wrong.}\\\mathrm{Hence,\;\;\lambda_1=0\;\;and\;\;\lambda_2=0}$

2. A triad of non-coplanar vector is linearly independent

If a, b, c are three non-zero, non-coplanar vectors and x, y, z are three scalars such that

$\quad \;\; x a+y b+z c =0 \\ \text { Then, } x=y=z=0$

Proof: It is given that xa + yb + zc = 0 ......(i)

Suppose that $x \neq 0$

Then Eq. (i) can be written as

$x =-y \mathbf{b}-z \mathbf{c} \\\\ \Rightarrow \mathbf{a} =-\frac{y}{x} \mathbf{b}-\frac{z}{x} \mathbf{c}\quad \ldots (ii)$

Now, $\frac{y}{x}$ and $\frac{z}{x}$ are scalars because x, y and are scalars. Thus, Eq. (ii) expresses a as a linear combination of b and c. Hence, a is coplanar with b and c which is contrary to our hypothesis because a, b and c are given to be non-coplanar. Thus, our supposition that $x \neq 0$ is wrong.

Hence, x=0.

Similarly, we can prove that y=0 and z=0.

Note: 4 vectors are always linearly dependent

Linear Dependent Vectors

$\\\mathrm{A\;system\;of\;vectors\;\;} \vec{a}_1, \;\vec{a}_2,\;\vec{a}_3 \dots\ldots \;, \vec{a}_n\text{ is said to be linearly dependent,\;if}\\\mathrm{there\;exists\;scalars\;\lambda_1,\;\lambda_2,\;\lambda_3,\ldots\ldots\ldots,\;\lambda_n\;not\;all\;zero\;such\;that}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\lambda_1\;\vec{a}_1+\lambda_2\;\vec{a}_2+\lambda_3\;\vec{a}_3+\ldots\ldots+\lambda_n\;\vec{a}_n=0$

It can be easily verified that

1. A pair of collinear vectors is linearly dependent.

2. A triad of coplanar vectors is linearly dependent.

Test of collinearity of three points

$Let\,\,A(\vec{a}), B(\vec{b})\,\,and\,\,C(\vec{c}), be\,\,three\,\,points\,\,in\,\,space$

$\\\text {If we can find x, y, z, not all zero such that }\\(i)\, x \vec{a}+y \vec{b}+z \vec{c}=0 \text { and (ii) } x+y+z=0$

Then the pointgs A, B and C are collinear

Proof:

$\\ \text {Let us suppose that points } A, B \text { and } C \text { are collinear and their position vectors are }\\ \vec{a},\; \vec{b}\; \text { and } \;\vec{c}\;\text { respectively. Let } C \text { divide the join of } \vec{a} \text { and } \vec{b} \text { in the ratio } y: x\\\text{Then,}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\vec{c}=\frac{x \vec{a}+y \vec{b}}{x+y}\\\mathrm{or\;\;\;\;\;\;\;\;\;\;\;\;\;\;} x \vec{a}+y \vec{b}-(x+y) \vec{c}=\overrightarrow{0}\\\mathrm{or\;\;\;\;\;\;\;\;\;\;\;\;\;\;} {x \vec{a}+y \vec{b}+z \vec{c}=\overrightarrow{0}, \text { where } z=-(x+y)} \\\mathrm{Also,\;\;\;\;\;\;\;\;\;} {x+y+z=x+y-(x+y)=0}\\$

Hence, the three point A, B and C are collinear.

Theorem 1:

$\\\text {If } \vec{a} \text { and } \vec{b} \text { are two non-zero, non-collinear vectors, then every vector } \vec{r} \text { coplanar with } \\\vec{a} \text { and } \vec{b}\;\text {can be expressed in one and only one way as a linear combination } \\ x \vec{a}+y \vec{b}\;\text{ where }x \;\text{and}\;y\;\text{being scalars.}$

Proof:

$\\\text {Let } O \text { be any point such that } \overrightarrow{O A}=\vec{a} \text { and } \overrightarrow{O B}=\vec{b}\\ {\text {As } \vec{r} \text { is coplanar with } \vec{a} \text { and } \vec{b}, \text { the lines } O A, O B \text { and } O R \text { are coplanar. }} \\ {\text {Through } R, \text { draw lines parallel to } O A \text { and } O B, \text { meeting them at } P \text { and } Q,} \\ {\text {respectively. Clearly, }}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\overrightarrow{O P}=x \overrightarrow{O A}=x \vec{a} \;\;\;\;\;\quad(\because \overrightarrow{O P} \text { and } \overrightarrow{O A} \text { are collinear vectors })\\\mathrm{Also,\;\;\;\;\;\;\;\;\;\;\;}\overrightarrow{O Q}=y \overrightarrow{O B}=y \vec{b}\;\;\;\;\;\;\quad(\because \overrightarrow{O Q} \text { and } \overrightarrow{O B}\text { are collinear vectors) }$

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\vec{r}=\overrightarrow{O R}=\overrightarrow{O P}+\overrightarrow{P R}=\overrightarrow{O P}+\overrightarrow{O Q} \quad(\because \overrightarrow{O Q} \text { and } \overrightarrow{P R} \text { are equal })\\ \mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=x \vec{a}+y \vec{b}\\\text { Thus, } \vec{r} \text { can be expressed in one way as a linear combination } x \vec{a}+y \vec{b}.$

Theorem 2

$\\\text {If }\; \vec{a}, \vec{b} \text { and } \vec{c} \text { are non-coplanar vectors, then any vector }\vec{r} \text { can be uniquely }\\\text{expressed as a linear combination }\;\vec{x} \vec{a}+y \vec{b}+z \vec{c} \;\text{ where} \;\;x, y \text { and } z \text { being scalars. }$

NOTE:

$\\\text{If vectors }\;\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k},\;\; \vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\;\;\text{and}\;\; \vec{c}=c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k} \\\text {are coplanar, then }\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\begin{vmatrix} a_1 &a_2 &a_3 \\ b_1 & b_2 & b_3\\ c_1 &c_2 & c_3 \end{vmatrix}=0$

(Proof of this will be seen in the concept of Scalar triple Product)

$\\\text {If vectors } x_{1} \vec{a}+y_{1} \vec{b}+z_{1} \vec{c}, \;\;x_{2} \vec{a}+y_{2} \vec{b}+z_{2} \vec{c} \;\text { and }\; x_{3} \vec{a}+y_{3} \vec{b}+z_{3} \vec{c} \;\;\text { are coplanar }\\\text{where }\;\vec{a}, \vec{b} \text { and } \vec{c} \text { are non-coplanar. Then, }\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;}\left|\begin{array}{lll}{x_{1}} & {y_{1}} & {z_{1}} \\ {x_{2}} & {y_{2}} & {z_{2}} \\ {x_{3}} & {y_{3}} & {z_{3}}\end{array}\right|=0$