Vector Algebra (Part-1) | S N Dey

Welcome to my website. This is this article, I am going to publish in this website about Vector chapter from S N dey mathematics class 12 or alternatively Chhaya mathematics. In this article, we are going to solve problems of Very Short Answer type questions from Exercise 1.

Unit -4 : VECTORS AND THREE DIMENSIONAL GEOMETRY [ Marks - 17 ]

1. Vectors [12 Periods ]

Vectors and scalars, magnitude and direction of a vector. Direction cosines/ratios of vectors. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector, addition of vectors, multi plication of a vector by a scalar, position vector of a point dividing a line segment in a given ratio. Scalar (dot) product of vectors, projection of a vector on a line. Vector (cross) product of vectors. Scalar triple product.

2. Three-Dimensional Geometry : [ 12 Periods ]

Direction cosines/ratios of a line joining two points. Cartesian and vector equation of a line, coplanar and skew lines, shortest distance between two lines. Cartesian and vector equation of a plane. Angle between (i) two lines, (ii) two planes, (iii) a line and a plane. Distance of a point from a plane.

Today we are going to solve Very Short Answer Type Questions of S N De Math book and will cover the exercise -1.

VSA TYPE QUESTIONS AND SOLUTIONS :

$~1(i)~$ Define a vector from geometrical concept.

Sol. A vector is an object that has both a magnitude and a direction. Geometrically, we can picture a vector as a directed line segment, whose length is the magnitude of the vector and with an arrow indicating the direction. The direction of the vector is from its tail to its head.

$~1(ii)~~$ Write two different vectors having the same magnitude.

Sol. Let $~\vec{a}~$ and $~\vec{b}~$ be two different vectors having the same magnitude which means $~~|\vec{a}|=|\vec{b}|.$

$~\text{Suppose}~\vec{a}= 2\hat{i}+\hat{j}+3\vec{k},~\vec{b}= 3\vec{i}+2\vec{j}-\vec{k}.$

$~\text{Then}~|\vec{a}|= \sqrt{2^2+1^2+3^2} = \sqrt{14}, \\~~~~~~ |\vec{b}|=\sqrt{3^2+2^2+(-1)^2}=\sqrt{14}~~\text{(ans.)}$

$~(iii)~~$ Write two different vectors having the same direction .

Sol. Two vectors say $~\vec{a},~\vec{b}~$ have the same direction if those two vectors are either collinear or parallel. There can be infinite number of such vectors having such properties as given in the problem.

$~\text{ let}~\vec{a}=\hat{i}+2\hat{j}+3\hat{k},~\vec{b}=2\hat{i}+4\hat{j}+6\hat{k}~\\~\text{so that}~\vec{b}=2(\hat{i}+2\hat{j}+3\hat{k})=2\vec{a} \rightarrow(1)$

So from $~(1)~$ , we can conclude that $~\vec{a}~$ and $~\vec{b}~$ are two different vectors having the same direction .

$~2.~$ Define the position vector of a point with respect to an origin. If $~\vec{a}~$ and $~\vec{b}~$ are the position vectors of the points $~P~$ and $~Q~$ respectively, find the vector $~\vec{PQ}~$ in terms of $~\vec{a}~$ and $~\vec{b}~.$

Let $~O~$ be any arbitrary point. Also let $~\vec{a}~$ and $~\vec{b}~$ be the position vectors of the points $~P~$ and $~Q~$ with respect to the point $~O~$ which is taken as origin. Then, $~\vec{OP}~$=$~\vec{a}~$ and $~\vec{OQ}~$=$~\vec{b}~$.

Now, using triangle law of addition in $~\Delta OPQ~$ we get,

$ \vec{OP}+\vec{PQ}=\vec{OQ} \\ \text{or,}~~ \vec{a}+\vec{PQ}=\vec{b} \\ \therefore~ \vec{PQ}=\vec{b}-\vec{a}.$

$~3.~$ Define a vector and a unit vector. Find in terms of vector $~\vec{a}~$ a unit vector in its direction .

Sol. A vector is a quantity that has both magnitude, as well as direction. A vector that has a magnitude of $~1~$ is called a unit vector.

If $~\vec{a}~$ is any vector, then a unit vector in its direction is $~\hat{a}=\frac{\vec{a}}{|\vec{a}|}.$

$~4.~$ If $~\vec{a}=2\hat{i}-5\hat{j}+3\hat{k}~$ and $~\vec{b}=\hat{i}-2\hat{j}-4\hat{k}~,$ find the value of $~|3\vec{a}+2\vec{b}|~.$

Solution.

$~3\vec{a}+2\vec{b} \\= 3(2\hat{i}-5\hat{j}+3\hat{k})+2(\hat{i}-2\hat{j}-4\hat{k}) \\= 6\hat{i}-15\hat{j}+9\hat{k}+2\hat{i}-4\hat{j}-8\hat{k} \\= 8\hat{i}-19\hat{j}+\hat{k}.\\ ~ \therefore~ |3\vec{a}+2\vec{b}|\\= \sqrt{8^2+(-19)^2+1^2} \\= \sqrt{64+361+1} \\= \sqrt{426}.$

$~5.~$ If $~\vec{a}=2\hat{i}+3\hat{j}-4\hat{k}~$ and $~\vec{b}=\hat{i}+2\hat{j}+\hat{k},~~$ then find $~(\vec{a}+\vec{b})~$ and $~|\vec{a}+\vec{b}|~.$

Solution.

$~\vec{a}+\vec{b}\\=(2\hat{i}+3\hat{j}-4\hat{k})+(\hat{i}+2\hat{j}+\hat{k})\\=(2\hat{i}+\hat{i})+(3\hat{j}+2\hat{j})+(-4\hat{k}+\hat{k})\\=3\hat{i}+5\hat{j}-3\hat{k} .\\~ \therefore~|\vec{a}+\vec{b}|\\=\sqrt{3^2+5^2+(-3)^2}\\=\sqrt{9+25+9}\\=\sqrt{43}$

$~6.~$ If $~\vec{\alpha}=2\hat{i}-5\hat{j}+4\hat{k}~$ and $~\vec{\beta}=\hat{i}-4\hat{j}+6\hat{k}~,~$ find $~(2\vec{\alpha}-\vec{\beta})~$. Also find a unit vector in the direction of the vector $~(2\vec{\alpha}-\vec{\beta})~$.

Solution.

$~2\vec{\alpha}-\vec{\beta} \\=2(2\hat{i}-5\hat{j}+4\hat{k})-(\hat{i}-4\hat{j}+6\hat{k}) \\ =4\hat{i}-10\hat{j}+8\hat{k}-\hat{i}+4\hat{j}-6\hat{k} \\= 3\hat{i}-6\hat{j}+2\hat{k}~.$

Now, a unit vector in the direction of the vector $~(2\vec{\alpha}-\vec{\beta})~$ is given by :

$~=\frac{2\vec{\alpha}-\vec{\beta}}{|2\vec{\alpha}-\vec{\beta}|}\\~~~= \frac{3\hat{i}-6\hat{j}+2\hat{k}}{|3\hat{i}-6\hat{j}+2\hat{k}|}\\~~~= \frac{3\hat{i}-6\hat{j}+2\hat{k}}{\sqrt{3^2+(-6)^2+2^2}}\\~~~= \frac{3\hat{i}-6\hat{j}+2\hat{k}}{\sqrt{9+36+4}}\\~~~= \frac{3\hat{i}-6\hat{j}+2\hat{k}}{\sqrt{49}}\\~~~= \frac 17(3\hat{i}-6\hat{j}+2\hat{k})~.$

$~7.~$ Show that the vectors $~ -\hat{i}+\hat{j},~-4\hat{i}-6\hat{j}~$ and $~5\hat{i}+5\hat{j}~$ are the sides of a right-angled triangle.

Solution.

$ |-\hat{i}+\hat{j}|^2\\=\left(\sqrt{(-1)^2+1^2}\right)^2\\=(\sqrt{1+1})^2\\=(\sqrt{2})^2\\=2 \rightarrow(1)$

$ |-4\hat{i}-6\hat{j}|^2\\=\left(\sqrt{(-4)^2+(-6)^2}\right)^2\\=(\sqrt{16+36})^2\\=(\sqrt{52})^2\\=52\rightarrow(2)$

$ |5\hat{i}+5\hat{j}|^2\\=\left(\sqrt{(5)^2+(5)^2}\right)^2\\=(\sqrt{25+25})^2\\=(\sqrt{50})^2\\=50\rightarrow(3)$

From $~(1),(2),(3)~$ we get,

$~~|-\hat{i}+\hat{j}|^2+ |5\hat{i}+5\hat{j}|^2=|-4\hat{i}-6\hat{j}|^2 \rightarrow(4)$

From $~(4),~$ we can conclude that the vectors $~ -\hat{i}+\hat{j},~-4\hat{i}-6\hat{j}~$ and $~5\hat{i}+5\hat{j}~$ are the sides of a right-angled triangle.

$~8.~$ The position vectors of the points $~A,B~$ and $~C~$ are $~-5\hat{i}+\hat{j},~5\hat{i}+5\hat{j}~$ and $~10\hat{i}+7\hat{j}~; $ Show that the points $~A,~B,~C~$ are collinear.

Sol. Let $~O~$ be the origin so that $~OA=-5\hat{i}+\hat{j},~OB=~5\hat{i}+5\hat{j},~OC=10\hat{i}+7\hat{j}.$

$~\text{Now,}~~\vec{AB}\\ = \vec{OB}-\vec{OA} \\ = (5\hat{i}+5\hat{j})- (-5\hat{i}+\hat{j}) \\ =5\hat{i}+5\hat{j}+5\hat{i}-\hat{j} \\= 10\hat{i}+4\hat{j}\rightarrow(1) $

$~\text{ Also,}~~ \vec{BC} \\ = \vec{OC}-\vec{OB} \\ =(10\hat{i}+7\hat{j})- (5\hat{i}+5\hat{j}) \\ =10\hat{i}+7\hat{j}-5\hat{i}-5\hat{j} \\ =5\hat{i}+2\hat{j}\rightarrow(2) $

So, from $~(1)~$ and $~(2)~$ we get, $~\vec{AB}=2\vec{BC} \rightarrow(3)$

Hence, from $~(3)~$ we can conclude that the points $~A,~B,~C~$ are collinear.

$~9.~$ Using vector method show that the points $~A~(-5,7),~B~(-4,5)~$ and $~C~(1,-5)~$ are collinear.

Sol. Let the position vectors of three points $~A,~B,~C~$ be

$~OA= -5\hat{i}+7\hat{j},~OB= -4\hat{i}+5\hat{j},~ OC= \hat{i}-5\hat{j},~$ where $~O~$ is the origin.

$~\text{Now,}~~\vec{AB}\\~~~~~~ = \vec{OB}-\vec{OA} \\ ~~~~~~= (-4\hat{i}+5\hat{j})- (-5\hat{i}+7\hat{j}) \\ ~~~~~~= -4\hat{i}+5\hat{j} +5\hat{i}-7\hat{j} \\ ~~~~~~=\hat{i}-2\hat{j}\rightarrow(1) $

$~\text{Also,}~~\vec{BC} \\~~~~~~ =\vec{OC}-\vec{OB} \\ ~~~~~~ =(\hat{i}-5\hat{j})-(-4\hat{i}+5\hat{j}) \\~~~~~~ =\hat{i}-5\hat{j}+4\hat{i}-5\hat{j} \\~~~~~~ = 5\hat{i}-10\hat{j} \\ ~~~~~~ = 5(\hat{i}-2\hat{j}) \\~~~~~~ =5\vec{AB}\rightarrow(2) $

Hence , from $~(1),~(2)~$ we can conclude that the points $~A,~B,~C~$ are collinear.

$~10.~$ If $~\vec{a}=\hat{i}+\hat{j}~$ and $~\vec{b}=4\hat{i}-\hat{j},~$ find a unit vector in the direction of the vector $~(2\vec{a}-\vec{b})~;$ Also find vector and scalar components of $~(2\vec{a}-\vec{b})~$ along two coordinate axes.

Solution.

$~2\vec{a}-\vec{b} \\= 2(\hat{i}+\hat{j})-(4\hat{i}-\hat{j}) \\ = 2\hat{i}+2\hat{j}-4\hat{i}+\hat{j} \\= -2\hat{i}+3j \rightarrow(1) $

a unit vector in the direction of the vector $~(2\vec{a}-\vec{b})~$ is given by :

$= \frac{2\vec{a}-\vec{b}}{|2\vec{a}-\vec{b}|} \\= \frac{-2\hat{i}+3j}{|-2\hat{i}+3j|} ~~~[\text{By (1)}] \\= \frac{-2\hat{i}+3j}{\sqrt{(-2)^2+3^2}} \\= \frac{1}{\sqrt{13}}(-2\hat{i}+3j).$

Now, vector component of $~(2\vec{a}-\vec{b})~$ along x-axis is : $~-2\hat{i}$

vector component of $~(2\vec{a}-\vec{b})~$ along y-axis is : $~3\hat{j}$

scalar component of $~(2\vec{a}-\vec{b})~$ along x-axis is : $~-2$

scalar component of $~(2\vec{a}-\vec{b})~$ along y-axis is : $~3.$

$~11.~$ The position vectors of two given points $~P~$ and $~Q~$ are $~8\hat{i}+3\hat{j}~$ and $~2\hat{i}-5\hat{j}~$ respectively ; find the magnitude and direction of the vector $~\vec{PQ}~.$

Sol. The position vectors of the points $~P~$ and $~Q~$ are given by :

$~\vec{OP}=8\hat{i}+3\hat{j},~~\vec{OQ}=2\hat{i}-5\hat{j}~,$ where $~O~$ is the origin.

$~\text{Now,}~\vec{PQ} \\~~~~= \vec{OQ}-\vec{OP} \\~~~~= (2\hat{i}-5\hat{j})-(8\hat{i}+3\hat{j}) \\~~~~= 2\hat{i}-5\hat{j}-8\hat{i}-3\hat{j} \\~~~~= -6\hat{i}-8\hat{j} \\ ~ \therefore~|\vec{PQ}|\\= \sqrt{(-6)^2+(-8)^2}=\sqrt{36+64}=\sqrt{100} \\~ \text{or,}~~|\vec{PQ}|=10. $

If $~\vec{PQ}~$ makes an angle $~\alpha~$ with $~\vec{OX}~$ (i.e. the positive direction of $~x~$-axis) , then

$~\cos\alpha \\= \frac{ \text{scalar component of}~\vec{PQ} ~\text{ along } ~\vec{OX}}{ |\vec{PQ}|} = \frac{-6}{10} \\ \therefore~\alpha = \cos^{-1} \left(-\frac 35\right) $

Therefore, required magnitude of $~\vec{PQ}~$ is $~10~$ and its direction makes an angle $~~\cos^{-1} \left(-\frac 35\right)~$ with $~~\vec{OX}~$ (i.e. the $~x~$-axis).

$~12.~$ If $~\vec{a}=4\hat{i}-3\hat{j}~$ and $~\vec{b}=-2\hat{i}+5\hat{j}~$ are the position vectors of the points and respectively ; find

$~(i)~$ the position vector of the middle point of the line-segment $~\overline{AB}~;$

$~(ii)~$ the position vectors of the points of trisection of the line segment $~\overline{AB}.$

Sol. $~(i)~$ The position vector of the middle point of the line-segment $~\overline{AB}~$ is given by :

$=\frac{\vec{a}+\vec{b}}{2} \\~~= \frac 12[(4\hat{i}-3\hat{j})+(-2\hat{i}+5\hat{j})] \\~~=\frac 12(4\hat{i}-3\hat{j}-2\hat{i}+5\hat{j}) \\~~= \frac 12(2\hat{i}+2\hat{j}) \\~~= \frac 12 \times 2(\hat{i}+\hat{j}) \\~~ = \hat{i}+\hat{j}~~\text{(ans.)}$

Sol. $~~(ii)~$

Let $~\vec{AB}~$ be trisected by $~C~$ in the ratio $~1:2~$ and trisected by $~D~$ in the ratio $~2:1$

$~\therefore~$ The position vector of $~C~$ is :

$=\frac{(-2\hat{i}+5\hat{j})+2(4\hat{i}-3\hat{j})}{1+2}\\=\frac{-2\hat{i}+5\hat{j}+8\hat{i}-6\hat{j}}{3}\\=\frac{6\hat{i}-\hat{j}}{3}\\=2\hat{i}-\frac 13\hat{j}.$

$~\therefore~$ The position vector of $~D~$ is :

$=\frac{2(-2\hat{i}+5\hat{j})+(4\hat{i}-3\hat{j})}{2+1}\\=\frac{-4\hat{i}+10\hat{j}+4\hat{i}-3\hat{j}}{3}\\=\frac{7\hat{j}}{3}\\=\frac 73 \hat{j}~~~\text{(ans.)}$

$~13.~$ Find the vector from the origin to the intersection of the medians of the triangle whose vertices are $~A(-1, -3), B(5, 7)~$ and $~C(2, 5).$

Sol. If $~\vec{\alpha}~$ denotes the vector from the origin to the intersection of the medians of the triangle with the given vertices, then coordinates of their intersection (which is also known as centroid of the given triangle ) is given by :

$~\left(\frac{-1+5+2}{3},\frac{-3+7+5}{3}\right) \\=\left(\frac 63,\frac 93\right)\\=(2,3)\rightarrow(1)$

Hence, $~~\vec{\alpha}=2\hat{i}+3\hat{j}~~[\text{ By (1)}]$

$~14(i).~$ Determine the value of $~p~$ for which the vectors $~p\hat{i}-5\hat{j}~$ and $~2\hat{i}-3\hat{j}~$ are collinear.

Sol. Two vectors will be collinear if

$~~p\hat{i}-5\hat{j} \\= m(2\hat{i}-3\hat{j})\\= (2m)\hat{i}-(3m)\hat{j}\rightarrow(1) $

Now, comparing both sides of $~(1)~$, we get,

$~~p=2m,~~5=3m \Rightarrow m =\frac 53 \\~~~~~~~~~~~ \therefore~~p = 2\times \frac 53 =\frac{10}{3}~~\text{(ans.)}$

$~(ii)~$ The sides $~AB~$ and $~BC~$ of the triangle $~ABC~$ are represented by the vectors $~2\hat{i}-\hat{j}+2\hat{k}~$ and $~\hat{i}+3\hat{j}+5\hat{k}~$ respectively ; find the vector representing side $~CA~$ of the triangle $~ABC$.

Solution.

$~\vec{AB}\\=2\hat{i}-\hat{j}+2\hat{k},~~\vec{BC}\\=\hat{i}+3\hat{j}+5\hat{k}.$

By triangle law of addition of vectors we get,

$ \vec{AC}\\=\vec{AB}+\vec{BC}\\=(2\hat{i}-\hat{j}+2\hat{k})+(\hat{i}+3\hat{j}+5\hat{k}) \\ \therefore ~\vec{AC}\\=(2+1)\hat{i}+(-1+3)\hat{j}+(2+5)\hat{k}\\=3\hat{i}+2\hat{j}+7\hat{k} \\ \therefore~~\vec{CA}\\=-\vec{AC}\\=-3\hat{i}-2\hat{j}-7\hat{k}~~\text{(ans.)}$

$~15.~$ The position vectors of the points $~A~$ and $~B~$ are $~3\hat{i}-\hat{j}+7\hat{k}~$ and $~4\hat{i}-3\hat{j}-\hat{k}~$; find the magnitude and direction cosines of $~\vec{AB}$.

Solution.

$~~~~ \vec{AB} \\= (\text{ position vector of }~B~)\\~~~- ( \text{ position vector of }~A~) \\=(4\hat{i}-3\hat{j}-\hat{k}) - (3\hat{i}-\hat{j}+7\hat{k}) \\ = 4\hat{i}-3\hat{j}-\hat{k}-3\hat{i}+\hat{j}-7\hat{k} \\ = \hat{i}-2\hat{j}-8\hat{k} \\ \therefore ~ |\vec{AB}|\\ = \sqrt{1^2+(-2)^2+(-8)^2} \\= \sqrt{1+4+64} \\= \sqrt{69} $

So, direction cosines of $~\vec{AB}~$ are $~~\frac{1}{\sqrt{69}},\frac{-2}{\sqrt{69}},\frac{-8}{\sqrt{69}}.$

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