NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.2 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.2.
Board | CBSE |
Textbook | NCERT |
Class | Class 12 |
Subject | Maths |
Chapter | Chapter 10 |
Chapter Name | Vector Algebra |
Exercise | Ex 10.2 |
Number of Questions Solved | 19 |
Category | NCERT Solutions |
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.2
Ex 10.2 Class 12 Maths Question 1.
Compute the magnitude of the following vectors:
\(\overrightarrow { a } =\hat { i } +\hat { j } +\hat { k } ,\overrightarrow { b } =\hat { 2i } -\hat { 7j } -\hat { 3k } \)
\(\overrightarrow { c } =\frac { 1 }{ \sqrt { 3 } } \hat { i } +\frac { 1 }{ \sqrt { 3 } } \hat { j } -\frac { 1 }{ \sqrt { 3 } } \hat { k } \)
Solution:
\(\overrightarrow { a } =\hat { i } +\hat { j } +\hat { k } \)
\(\left| \overrightarrow { a } \right| =\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } } \)
Ex 10.2 Class 12 Maths Question 2.
Write two different vectors having same magnitude.
Solution:
\(\overrightarrow { a } =\hat { i } +\hat { 2j } +\hat { 3k } ,\overrightarrow { b } =\hat { 3i } +\hat { 2j } +\hat { k } \)
Such possible answers are infinite
Ex 10.2 Class 12 Maths Question 3.
Write two different vectors having same direction.
Solution:
Let the two vectors be
\(\overrightarrow { a } =\hat { i } +\hat { j } +\hat { k } ,\overrightarrow { b } =\hat { 3i } +\hat { 3j } +\hat { 3k } \)
Hence vectors \(\overrightarrow { a } ,\overrightarrow { b } \) have the same direction but different magnitude
Ex 10.2 Class 12 Maths Question 4.
Find the values of x and y so that the vectors \(\overrightarrow { 2i } +\overrightarrow { 3j } \quad and\quad \hat { xi } +\hat { yj } \) are equal.
Solution:
We are given \(\overrightarrow { 2i } +\overrightarrow { 3j } \quad and\quad \hat { xi } +\hat { yj } \)
If vectors are equal, then their respective components are equal. Hence x = 2, y = 3.
Ex 10.2 Class 12 Maths Question 5.
Find the scalar and vector components of the vector with initial point (2,1) and terminal point (-5,7).
Solution:
LetA(2, 1) be the initial point and B(-5,7) be the terminal point \(\overrightarrow { AB } =\left( { x }_{ 2 }-{ x }_{ 1 } \right) \hat { i } +\left( { y }_{ 2 }-{ y }_{ 1 } \right) \hat { j } =-\hat { 7i } +\hat { 6j } \)
∴The vector components are \(-\hat { 7i } and\hat { 6j } \) and scalar components are – 7 and 6.
Ex 10.2 Class 12 Maths Question 6.
Find the sum of three vectors:
\(\overrightarrow { a } =\hat { i } -\hat { 2j } +\hat { k } ,\overrightarrow { b } =-2\hat { i } +\hat { 4j } +5\hat { k } \quad and\quad \overrightarrow { c } =\hat { i } -\hat { 6j } -\hat { 7k } ,\)
Solution:
\(\overrightarrow { a } =\hat { i } -\hat { 2j } +\hat { k } ,\overrightarrow { b } =-2\hat { i } +\hat { 4j } +5\hat { k } \quad and\quad \overrightarrow { c } =\hat { i } -\hat { 6j } -\hat { 7k } ,\)
\(\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =\hat { 0i } -\hat { 4j } -\hat { k } =-4\hat { i } -\hat { k } \)
Ex 10.2 Class 12 Maths Question 7.
Find the unit vector in the direction of the vector
\(\overrightarrow { a } =\hat { i } +\hat { j } +\hat { 2k } \)
Solution:
\(\overrightarrow { a } =\hat { i } +\hat { j } +\hat { 2k } \)
Ex 10.2 Class 12 Maths Question 8.
Find the unit vector in the direction of vector \(\overrightarrow { PQ } \), where P and Q are the points (1,2,3) and (4,5,6) respectively.
Solution:
The points P and Q are (1, 2, 3) and (4, 5, 6) respectively
\(\overrightarrow { PQ } =(4-1)\hat { i } +(5-2)\hat { j } +(6-3)\hat { k } \)
Ex 10.2 Class 12 Maths Question 9.
For given vectors \(\overrightarrow { a } =2\hat { i } -\hat { j } +2\hat { k } \quad and\quad \overrightarrow { b } =-\hat { i } +\hat { j } -\hat { k } \) find the unit vector in the direction of the vector \(\overrightarrow { a } +\overrightarrow { b } \)
Solution:
\(\overrightarrow { a } =2\hat { i } -\hat { j } +2\hat { k } \quad and\quad \overrightarrow { b } =-\hat { i } +\hat { j } -\hat { k } \)
Ex 10.2 Class 12 Maths Question 10.
Find a vector in the direction of \(5\hat { i } -\hat { j } +2\hat { k } \) which has magnitude 8 units.
Solution:
The given vector is \(\overrightarrow { a } =5\hat { i } -\hat { j } +2\hat { k } \)
Ex 10.2 Class 12 Maths Question 11.
Show that the vector \(2\hat { i } -3\hat { j } +4\hat { k } \quad and\quad -4\hat { i } +6\hat { j } -8\hat { k } \) are collinear.
Solution:
\(\overrightarrow { a } =2\hat { i } -3\hat { j } +4\hat { k } \quad and\quad \overrightarrow { b } =-4\hat { i } +6\hat { j } -8\hat { k } \)
\(=-2(2\hat { i } -3\hat { j } +4\hat { k } ) \)
vector \(\overrightarrow { a } \quad and\quad \overrightarrow { b } \) have the same direction they are collinear.
Ex 10.2 Class 12 Maths Question 12.
Find the direction cosines of the vector \(\hat { i } +2\hat { j } +3\hat { k } \)
Solution:
let \(\overrightarrow { p } =\hat { i } +2\hat { j } +3\hat { k } \)
Now a = 1,b = 2,c = 3
Ex 10.2 Class 12 Maths Question 13.
Find the direction cosines of the vector joining the points A (1,2, -3) and B(-1, -2,1), directed fromAtoB.
Solution:
Vector joining the points A and B is
\(({ x }_{ 2 }-{ x }_{ 1 })\hat { i } +({ y }_{ 2 }-{ y }_{ 1 })\hat { j } +({ z }_{ 2 }-{ z }_{ 1 })\hat { k } \)
Ex 10.2 Class 12 Maths Question 14.
Show that the vector \(\hat { i } +\hat { j } +\hat { k } \) are equally inclined to the axes OX, OY, OZ.
Solution:
Let \(\hat { i } +\hat { j } +\hat { k } =\overrightarrow { a } \) , Direction cosines of vector \(x\hat { i } +y\hat { j } +z\hat { k } \) are
which shows that the vector a is equally inclined to the axes OX, OY, OZ.
Ex 10.2 Class 12 Maths Question 15.
Find the position vector of a point R which divides the line joining the points whose positive vector are \(P(\hat { i } +2\hat { j } -\hat { k } )\quad and\quad Q(-\hat { i } +\hat { j } +\hat { k } )\) in the ratio 2:1
(i) internally
(ii) externally.
Solution:
(i) The point R which divides the line joining the point \(P(\overrightarrow { a } )\quad and\quad Q(\overrightarrow { b } )\) in the ratio m : n
Ex 10.2 Class 12 Maths Question 16.
Find position vector of the mid point of the vector joining the points P (2,3,4) and Q (4,1, -2).
Solution:
Let \(\overrightarrow { OP } =2\hat { i } +3\hat { j } +4\hat { k } \quad and\quad \overrightarrow { OQ } =4\hat { i } +\hat { j } -2\hat { k } \)
Ex 10.2 Class 12 Maths Question 17.
Show that the points A, B and C with position vector \(\overrightarrow { a } =3\hat { i } -4\hat { j } -4\hat { k } ,\overrightarrow { b } =2\hat { i } -\hat { j } +\hat { k } and\quad \overrightarrow { c } =\hat { i } -3\hat { j } -5\hat { k } \) respectively form the vertices of a right angled triangle.
Solution:
\(\overrightarrow { AB } =\overrightarrow { b } -\overrightarrow { a } =-\hat { i } +3\hat { j } +5\hat { k } \)
Ex 10.2 Class 12 Maths Question 18.
In triangle ABC (fig.), which of the following is not
(a) \(\overrightarrow { AB } +\overrightarrow { BC } +\overrightarrow { CA } =\overrightarrow { 0 } \)
(b) \(\overrightarrow { AB } +\overrightarrow { BC } -\overrightarrow { AC } =\overrightarrow { 0 } \)
(c) \(\overrightarrow { AB } +\overrightarrow { BC } -\overrightarrow { CA } =\overrightarrow { 0 } \)
(d) \(\overrightarrow { AB } -\overrightarrow { CB } +\overrightarrow { CA } =\overrightarrow { 0 } \)
Solution:
We know that
\(\overrightarrow { AB } +\overrightarrow { BC } +\overrightarrow { CA } =\overrightarrow { 0 } \)
\(\overrightarrow { AB } +\overrightarrow { BC } -\overrightarrow { AC } =\overrightarrow { 0 } \)
Hence option (c) is not correct
Ex 10.2 Class 12 Maths Question 19.
If \(\overrightarrow { a } ,\overrightarrow { b } \) are two collinear vectors then which of the following are incorrect:
(a) \(\overrightarrow { b } =\lambda \overrightarrow { a } \), for some scalar λ.
(b) \(\overrightarrow { a } =\pm \overrightarrow { b } \)
(c) the respective components of \(\overrightarrow { a } ,\overrightarrow { b } \) are proportional.
(d) both the vectors \(\overrightarrow { a } ,\overrightarrow { b } \) have same direction, but different magnitudes.
Solution:
Options (d) is incorrect since both the vectors \(\overrightarrow { a } ,\overrightarrow { b } \) , being collinear, are not necessarily in the same direction. They may have opposite directions. Their magnitudes may be different.
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