Algebra of Vectors Excercise:22.5
Algebra of vector exercise 22.5 question 1
Answer:
5
Hints:
$\left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}=\left | \vec{a} \right |$Given: Position vector of a point (-4,-3) be
$\left | \vec{a} \right |$Find:
$\left | \vec{a} \right |$We know position vector of a point(x,y) is given by
$x\hat{i} +y\hat{j}$ where
$\hat{i}$ and
$\hat{j}$ are units vectors in x and y directions.
$\vec{a}=(-4)\hat{i} +(-3)\hat{j}$Now we need to find magnitude of
$\vec{a}$ i.e
$\left |\vec{a} \right |$$\left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}$Here, x=-4 and y=-3
$\left |\vec{a} \right |=\sqrt{(-4)^{2} +(-3)^{2}}\\ \left |\vec{a} \right |=\sqrt{16+9}\\ \left |\vec{a} \right |=\sqrt{25}\\ \left |\vec{a} \right |=5$thus
$\left |\vec{a} \right |=5$Algebra of vector exercise 22.5 question 2
Answer:
$n=\pm 5$Hints:
$\left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}=\left | \vec{a} \right |$Given: Position vector of
$\vec{a}$ point (12,n) be
$\left | \vec{a} \right |$Find:
$\left | \vec{a} \right |$We know position vector of a point(x,y) is given by
$x\hat{i} +y\hat{j}$ where
$\hat{i}$ and
$\hat{j}$ are units vectors in x and y directions.
$\vec{a}=12\hat{i} +n\hat{j}$Then:
$\left | \vec{a} \right |=\sqrt{x^{2}+y^{2}}$Here, x=12 and y=n
$\left |\vec{a} \right |=\sqrt{12^{2} +n^{2}}\\ \left |\vec{a} \right |=13$[given]
$13=\sqrt{(12)^{2} +(n)^{2}}\\$Squaring both sides
$13^{2}=(12)^{2} +(n)^{2}\\ 169=144+n^{2}\\ n^{2}=25\\ n=\pm 5$Algebra of vector exercise 22.5 question 4 (i)
Answer:
$\vec{AB}=-3\hat{i}+4\hat{j}\\ \left |\vec{AB} \right |=5$Hint:
$\left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}=\left | \vec{a} \right |$Position vector
$\left |\vec{AB} \right |$ is given x, y
$\left |\vec{AB} \right |$ = position vector of B - position vector of A
Given:
A=(4,-1) and B(1,3)
Solution:
We know position vector of a point (x, y) is given by
$x\hat{i} +y\hat{j}$ where
$\hat{i}$ and
$\hat{j}$ are unit vector in
x and y direction
$A=4\hat{i}+(-1)\hat{j}\\ B=i+3\hat{j}$Then the position vector
$\vec{AB}$ is given by
$\vec{AB}$ = position vector of B - position vector of A
$=\left (\hat{i}+3\hat{j} \right )-\left (4\hat{i}-\hat{j} \right )\\ =\hat{i}+3\hat{j} -4\hat{i}+\hat{j} \\ AB=-3\hat{i}+4j$So,
$\left |\vec{AB} \right |=(-3)^{2}+(4)^{2}\\ =\sqrt{9+16}=\sqrt{25}\\\left |\vec{AB} \right |=5$Algebra of vector exercise 22.5 question 4 (ii)
Answer:
$\vec{AB}=4\hat{i}-8\hat{j}\\ \left |\vec{AB} \right |=4\sqrt{5}$Hint:
$\left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}=\left | \vec{a} \right |$Position vector
$\vec{AB}$ is given x, y
$\vec{AB}$ = position vector of B - position vector of A
Given:
A=(-6,3) and B(-2,-5)
Solution:
We know position vector of a point (x, y) is given by
$x\hat{i} +y\hat{j}$ where
$\hat{i}$ and
$\hat{j}$ are unit vector in
x and y direction
$\vec{A}=-6\hat{i}+3\hat{j}\\ \vec{B}=-2i-5\hat{j}$Then the position vector
$\vec{AB}$ is given by
$\vec{AB}$ = position vector of B - position vector of A
$=\left (2\hat{i}-5\hat{j} \right )-\left (-6\hat{i}+3\hat{j} \right )\\ =-2\hat{i}-5\hat{j} +6\hat{i}-3\hat{j} \\ =4\hat{i}-8\hat{j}\\ \left |\vec{AB} \right |=\sqrt{4^{2}+(-8)^{2}}=\sqrt{16+64}\\ =\sqrt{80}\\ \left |\vec{AB} \right |=4\sqrt{5}$-
Algebra of Vectors exercise 22.5 question 5
Answer:
(-1,-2)
Hint: Position vector
$\vec{AB}$ is given
$\vec{AB}$ = position vector B- position vector of A
Given: Coordinate of A (-1, 3)
Coordinates of B (-2,1)
Solution:
$\vec{A}= -\hat{i}+3\hat{j}$ &
$\vec{B}= -2\hat{i}+\hat{j}$We know position vector of point (x, y) is given by
$(x\hat{i}+y\hat{j})$Let 0 be the origin
Let P(x,y) be the second point
Then
$\vec{P}$ is the tip of the position vector
$\vec{OP}$ of the point p
We have:
$\vec{OP}=x\hat{i}+y\hat{j}$position vector of B- position vector of A
$\vec{AB}$= position vector of B= position vector of A
$=(-2\hat{i}+\hat{j})-(-\hat{i}+3\hat{j})\\ =-2\hat{i}+\hat{j}+\hat{i}-3\hat{j}\\ =-\hat{i}-2\hat{j}$$\vec{OP}=\vec{AB}$ given
$x\hat{i}+y\hat{j}=-\hat{i}-2\hat{j}$then x=-1 and y= -2
Hence coordinate of the required point is (-1,-2)
Algebra of Vectors exercise 22.5 question 6
Answer:
the coordinates of D is (-4,-3)
Hint: Position vectors
$\vec{AB}$ is given by = position vector of B -position vector of A
Given;
Coordinates of A=(-2,-1)
Coordinates of B=(3,0)
Coordinates of C=(1,-2)
Let the coordinates of D is (x , y)
Since ABCD is a parallelogram,
AB=DC
We have
$\vec{AB}=\vec{DC}$As we know position vector of point (x, y) is given by
$x\hat{i}=y\hat{j}$where
$\hat{i}$ and
$\hat{j}$ are position vector of x, and y direction.
$\vec{A}=-2\hat{i}-\hat{j}\\ \vec{B}=3\hat{i}+0\hat{j}\\ \vec{C}=\hat{i}-2\hat{j}\\ \vec{D}=x\hat{i}+y\hat{j}$Then the position vector
$\vec{AB}$ is given by
$\vec{AB}$= position vector of B - position vector of A
$\left (3\hat{i}+0\hat{j} \right )-\left (-2\hat{i}-\hat{j} \right )\\ 3\hat{i}+0\hat{j} +2\hat{i}+\hat{j} \\ \vec{AB}=5\hat{i}+\hat{j}$Then we find
$\vec{DC}$$\vec{DC}$= position vector C- position vector of D
$=\left (\hat{i}-2\hat{j} \right )-\left (x\hat{i}+y\hat{j} \right )\\ =\hat{i}-x\hat{i}-2\hat{j}+y\hat{j} \\ =\hat{i}\left ( 1-x \right )+\hat{j}\left ( -2-y \right )$We have
$\vec{AB}=\vec{DC}$Then:
$5\hat{i}+\hat{j}=\left ( 1-x \right )\hat{i}+\left ( -2-y \right )\hat{j}$Then comparing coefficient of
$\hat{i}$ and
$\hat{j}$$\begin{matrix} 5=1-x & 1=-2-y\\ x=1-5 &y=-2-1 \\ x=-4 & y=-3 \end{matrix}$Hence the coordinates of D is (-4, -3)
Algebra of Vectors exercise 22.5 question 7
Answer:
$\hat{i}-5\hat{j}$Hint: position vector of point (x , y) is given by where
$\hat{i}$ and
$\hat{j}$ are the position in x an y direction.
Given:
Point
$A(3,4),B(5,-6),C(4,-1)$Find
$\vec{a}+2\vec{b}-3\vec{c}$Solution:
As we know position vector of point (x, y) is given by
$\left (x\hat{i}+y\hat{j} \right )$Where
$\hat{i}$and
$\hat{j}$ are the position vector in x and y direction
Then,
$\vec{A}=\left (3\hat{i}+4\hat{j} \right )\\ \vec{B}=\left (5\hat{i}+(-6)\hat{j} \right )=5\hat{i}-6\hat{j}\\ \vec{C}=\left (4\hat{i}+(-1)\hat{j} \right )=4\hat{i}-\hat{j}\\ 2\vec{B}=2\left (5\hat{i}-6\hat{j} \right )=10\hat{i}-12\hat{j}\\ 3\vec{C}=3\left (4\hat{i}-\hat{j} \right )=12\hat{i}-3\hat{j}$$\vec{a}+2\vec{b}-3\vec{c}$ Substitute value of
$\vec{a},2\vec{b},3\vec{c}$ in equation
$\vec{a}+2\vec{b}-3\vec{c}$$=3\hat{i}+4\hat{j}+10\hat{i}-12\hat{j}-\left ( 12\hat{i}-3\hat{j} \right )\\ =3\hat{i}+4\hat{j}+10\hat{i}-12\hat{j}- 12\hat{i}+3\hat{j}\\ =\hat{i}\left ( 3+10-12 \right )+\hat{j}\left ( 4-12+3 \right )\\ =\hat{i}\left ( 1 \right )+\hat{j}\left ( -5 \right )\\ =\hat{i}-5\hat{j}$Algebra of Vectors exercise 22.5 question 8
Answer:
(9, -4)
Hint: position vector
$\vec{AB}$ is given by
$\vec{AB}$ =position vector of B- position vector of A
Given: coordinates of A(4, -1) , a
$\vec{a}$ is the position vector whose tip i is (5, -3)
Solution:
Let O be the origin of point (0, 0)
Position P(5, -3) be the tip of the position vector
$\vec{a}$Then the
$\vec{a}$ ,
$\vec{OP}$ =Position vector of p- position vector of o.
As we know position vector of point (x, y) is given by
$\left ( x\hat{i} + y\hat{j}\right )$ where
$\hat{i}$ and
$\hat{j}$ are the unit vectors.
Then
$\vec{P}=5\hat{i}+-3\hat{j}=5\hat{i}-3\hat{j}\\ \vec{O}=0\hat{i}+0\hat{j} \\ \vec{OP}=5\hat{i}-3\hat{j}-0\hat{i}+0\hat{j}\\ \vec{OP}=5\hat{i}-3\hat{j}-0\hat{i}-0\hat{j}\\ =5-0\hat{i}+-3-0\hat{j}\\ \vec{OP}=5\hat{i}-3\hat{j}$Let the coordinate of B be (x,y) and A has coordinate (4,-1)
Therefore
$\vec{A}=4\hat{i}+-1\hat{j}=4\hat{i}-\hat{j}\\ \vec{B}=x\hat{i}+y\hat{j}$AB=
$\vec{AB}$ Position vectors of B – position vector of A
$\vec{AB}=x\hat{i}+y\hat{j}-\left (4\hat{i}-\hat{j} \right )\\ =x\hat{i}+y\hat{j}-4\hat{i}+\hat{j}\\ =\left (x-4 \right )\hat{i}+\hat{j}\left (y+1 \right )$Now,
$\vec{AB}=\vec{a}\\ \left ( x-4 \right )\hat{i}+\left ( y+1 \right )\hat{j}=5\hat{i}-3\hat{j}$Comparing coefficient of
$\hat{i}$ &
$\hat{j}$$x-4=5\\ x=5+4\\ x=9\\ x+1=-3\\ y=-3-1\\ y=-4$Hence, the coordinate of B are (9,-4)
Algebra of Vector exercise 22.5 question 9
Answer:
the magnitude of AB and AC is equal
Hence the points
$2\hat{i},-\hat{i}-4\hat{j}$ and
$-\hat{i}+4\hat{j}$ form an isosceles triangle
Hint: the magnitude of
$\left |\vec{a} \right |=\sqrt{x^{2}+y^{2}}$Position vector
$\vec{AB}$ is given by the
$\vec{AB}$ = position vector of B - position vector of A
Given: the points A,B,C with position vectors
$\vec{a},\vec{b},\vec{c}$ respectively
Also,
$\vec{a}=2\hat{i}\\ \vec{b}=-\hat{i}-4\hat{j}\\ \vec{c}=-\hat{i}+4\hat{j}$Solution:Now we find the position vector
$\vec{AB}$Then,
$\vec{AB}=\vec{b}-\vec{a}\\ =\left (-\hat{i}-4\hat{j} \right )-2\hat{i}\\ =-\hat{i}-4\hat{j}-2\hat{i}\\ \vec{AB}=-3\hat{3}-4\hat{j}$Now
$\left |\vec{AB} \right |=\sqrt{x^{2}+y^{2}}$Here x=-3 and y= -4
$=\sqrt{(-3)^{2}+(-4)^{2}}\\ =\sqrt{9+16}=\sqrt{25}\\ =5$$\vec{BC}$=Position vector of C – Position vector of B
$=\left ( -\hat{i}+4\hat{j} \right )-\left ( -\hat{i}-4\hat{j} \right )\\ = -\hat{i}+4\hat{j}+\hat{i}+4\hat{j} \\ =0\hat{i}+8\hat{j}\\ =8\hat{j}$$\vec{AC}$=Position vector of C – position vector of A
$=\left ( -\hat{i}+4\hat{j} \right )-\left ( 2\hat{i} \right )\\ = -\hat{i}+4\hat{j}-\hat{j} \\ =-3\hat{i}+4\hat{j}\\$Now
$\left |\vec{AC} \right |=\sqrt{x^{2}+y^{2}}$Here x=-3 and y=4
$=\sqrt{(-3)^{2}+(-4)^{2}}\\ =\sqrt{9+16}=\sqrt{25}\\ =5$Since, the magnitude of AB and AC are equal
Hence, the points
$2\hat{i},-\hat{i}-4\hat{j}$ and
$-\hat{i}+4\hat{j}$ form an isosceles triangle
Algebra of Vector exercise 22.5 question 10
Answer:
$\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}$Hint: a unit vector parallel to
$\vec{a}=\hat{a}=\frac{\vec{a}}{\left | \vec{a} \right |}\\ \left | \vec{a} \right |=\sqrt{x^{2}+y^{2}}$Given:
$\vec{a}=\hat{i}+\sqrt{3}\hat{j}$Solution:
We can find the magnitude of
$\vec{a}$$\left | \vec{a} \right |=\sqrt{x^{2}+y^{2}}$Here x = 1 and
$y=\sqrt{3}$$\left |\vec{a} \right |=1^{2}+(\sqrt{3})^{2}\\ =\sqrt{1+3}=4\\ \left |\vec{a} \right |=2$Unit vector parallel to
$\vec{a}=\hat{a}=\frac{\vec{a}}{\left | \vec{a} \right |}\\$$=\frac{1}{2}\left ( \hat{i}+\sqrt{3}\hat{j} \right )\\ =\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}$Algebra of Vector exercise 22.5 question 11
Answer:
$\lambda=8$ and
$\mu =-5$Hint: here we use sector formula
$\frac{\left ( mx_{2}+nx_{1} \right )}{m+n},\frac{\left ( my_{2}+ny_{1} \right )}{m+n}$Where m,n are the given ratios
Given: the position vectors of points A, B and C are
$\vec{A}=\lambda \hat{i}+3\hat{j}\\\vec{B}=12\hat{i}+\mu \hat{j}\\\vec{C}=11\hat{i}-3\hat{j}$Solution:
It is given that C, divides the line segment joining A and B in the ratio 3:1
$\begin{aligned} &11 \hat{\imath}-3 \hat{\jmath}=3 \times(12 \hat{\imath}+\mu \hat{\jmath})+1(\lambda \hat{\imath}+3 \hat{\jmath}) \\ &11 \hat{\imath}-3 \hat{\jmath}=\frac{36 \hat{\imath}+3 u \hat{\jmath}+\lambda \hat{\imath}+3 \hat{\jmath}}{4} \\ \end{aligned}$$\begin{aligned}&11 \hat{\imath}-3 \hat{\jmath}=\frac{(36+\lambda) \hat{\imath}+(3 \mu+3) \hat{\jmath}}{4} \\ &4(11 \hat{\imath}-3 \hat{\jmath})=(36+\lambda) \hat{\imath}+(3 u+3) \hat{\jmath} \\ &44 \hat{\imath}-12 \hat{\jmath}=(36+\lambda) \hat{\imath}+(3 u+3) \hat{\jmath} \end{aligned}$Equating the corresponding components, we get;
$36+\lambda=44\\ \lambda=44-36\\ \lambda=8$and
$3\mu+3=-12\\ 3\mu=-12-3=-15\\ \mu=\frac{-15}{3}=-5\\ \mu=-5$Thus, the values of
$\lambda$ and
$\mu$ are 8 and -5 respectively
Algebra of vectors exercise 22.5, question 12 (i)
Answer:
: The components of
$\vec{P}$ along x axis is a vector of magnitude 3 along the positive direction of x axis
The component of
$\vec{P}$ along y axis is a vector of magnitude 2 along the positive direction of Y axis
$3\hat{i},2\hat{j}$Hint: position vector of point (x, y) is given by
$x\hat{i}+y\hat{j}$ where
$\hat{i}$ and
$\hat{j}$ are the unit vector in x and y direction
Given: P(3,2)
Solution:
As we know, position vector of a point (x, y) is given by
$x\hat{i}+y\hat{j}$Then,
Position vector of a point (3, 2) is
$\vec{p}=3\hat{i}+2\hat{j}$The components of
$\vec{P}$along x axis is a vector of magnitude 3 along the positive direction of x axis
The component of
$\vec{P}$ along y axis is a vector of magnitude 2 along the positive direction of Y axis
Algebra of vectors exercise 22.5, question 12 (ii)
Answer:
The components of
$\vec{Q}$ along x axis is a vector of magnitude 5 along the negative direction of x axis
The component of
$\vec{Q}$ along y axis is a vector of magnitude 1 along the positive direction of Y axis
$-5\hat{i},\hat{j}$Hint: position vector of point (x, y) is given by
$x\hat{i}+y\hat{j}$ where
$\hat{i}$ and
$\hat{j}$ are the unit vector in x and y direction
Given:Q=(-5,1)
Solution:
As we know, position vector of a point (x, y) is given by
$x\hat{i}+y\hat{j}$So
$\vec{Q}=-5\hat{i}+1\hat{j}\\ \vec{Q}=-5\hat{i}+\hat{j}\\$The components of
$\vec{Q}$ along x axis is a vector of magnitude 5 along the negative direction of x axis
The component of
$\vec{Q}$ along y axis is a vector of magnitude 1 along the positive direction of Y axis
Algebra of vectors exercise 22.5, question 12 (iii)
Answer:
The components of
$\vec{R}$ along x axis is a vector of magnitude 11 along the negative direction of x axis
The component of
$\vec{R}$ along y axis is a vector of magnitude 9 along the negative direction of Y axis
$-11\hat{i},-9\hat{j}$Hint: position vector of point (x, y) is given by
$x\hat{i}+y\hat{j}$ where
$\hat{i}$ and
$\hat{j}$ are the unit vector in x and y direction
Given:R=(-11,-9)
Solution:
As we know, position vector of a point (x, y) is given by
$x\hat{i}+y\hat{j}$So
$\vec{R}=-11\hat{i}+(-9)\hat{j}\\ \vec{R}=-11\hat{i}-9\hat{j}\\$The components of
$\vec{R}$ along x-axis is a vector of magnitude 11 along the negative direction of x axis
The component of
$\vec{R}$ along y-axis is a vector of magnitude 9 along the negative direction of Y axis
Algebra of vectors exercise 22.5, question 12 (iv)
Answer:
The components of
$\vec{S}$ along x axis is a vector of magnitude 4 along the positive direction of x axis
The component of
$\vec{S}$ along y axis is a vector of magnitude 3 along the negative direction of Y axis
$4\hat{i},-3\hat{j}$Hint: position vector of point (x, y) is given by
$x\hat{i}+y\hat{j}$ where
$\hat{i}$ and
$\hat{j}$ are the unit vector in x and y direction
Given:S=(-4,-3)
Solution:
As we know, position vector of a point (x, y) is given by
$x\hat{i}+y\hat{j}$So
$\vec{S}=4\hat{i}+(-3)\hat{j}\\ \vec{S}=4\hat{i}-3\hat{j}\\$The components of
$\vec{S}$ along x-axis is a vector of magnitude 4 along the positive direction of x-axis
The component of
$\vec{S}$ along y-axis is a vector of magnitude 3 along the negative direction of Y-axis
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