Class 12 RD Sharma Solutions – Chapter 23 Algebra of Vectors – Exercise 23.2
Last Updated :
08 May, 2021
Question 1. If P, Q, and R are three collinear points such that and . Find the vector
Solution:
According to the question, given that
Points P, Q, and R are collinear.
Also, and
So,
Question 2. Given condition that three vectors and form the three sides of a triangle. What are other possibilities?
Solution:
According to the question, given that are three sides of a triangle ABC.
[since ]
[since ]
So,
As we know that if vectors are represented in magnitude and direction by the two sides
of triangle taken is same order, then their sum is represented by the third side taken in reverse order.
So,
or
Question 3. If and are two non- collinear vectors having the same initial point. What are the vectors represented by and ?
Solution:
According to the question, given that and
are two non-collinear vectors having the same initial point.
So, let us considered
Now we draw a parallelogram named as ABCD
Using the properties of parallelogram, we get
In ∆ABC,
Using the triangle law, we get
…….(i)
In ∆ABD,
Using the triangle law, we get
…….(ii)
On solving equation (i) and (ii), we get
and
are diagonals of a parallelogram whose adjacent sides are and
Question 4. If is a vector and m is a scalar such that , then what are the alternatives for m and ?
Solution:
According to the question, given that m is a scalar and is a vector such that
[since let ]
Now on comparing the coefficients of of LHS and RHS, we get
ma1 = 0 ⇒ m = 0 or a1 = 0 …….(i)
mb1 = 0 ⇒ m = 0 or b1 = 0 …….(ii)
mc1 = 0 ⇒ m = 0 or c1 = 0 …….(iii)
Now from eq (i), (ii) and (iii), we get
m = 0 or a1 = b1 = c1 = 0
m = 0 or
m = 0 or
Question 5. If are two vectors, then write the truth value of the following statement:
(i)⇒
(ii)
(iii)
Solution:
(i) Let us assume
Given that, a = -b
So,
Now on comparing the coefficients of i, j, k in LHS and RHS, we get
a1 = a2 …….(i)
b1 = b2 …….(ii)
c1 = c2 …….(iii)
From eq(i), (ii), and (iii),
(ii) Given a and b are two vectors such that
So, it means the magnitude of vector is equal to the magnitude
of vector , but we cannot find the direction of the vector.
Hence, it is false that
(iii) Given for any vector
are equal but we cannot find the direction of the vector of
So, it is false.
Question 6. ABCD is a quadrilateral. Find the sum of the vectors and .
Solution:
According to the question,
ABCD is a quadrilateral.
so,
In ∆ADC,
By using triangle law, we get
……(i)
In ∆ABC,
By using triangle law, we get
……(ii)
Now put the value of in equation (ii), we get
Now on addingon both sides,
Question 7. ABCDE is a pentagon, prove that
(i)
(ii)
Solution:
(i) According to the question,
ABCDE is a pentagon,
So,
Using the law of triangle , we get
Using triangle law ,, we get
= 0
Hence Proved
(ii) According to the question,
ABCDE is a pentagon,
So,
Using triangle law,, we get
Hence Proved
Question 8. Prove that the sum of all vectors drawn from the centre of a regular octagon to its vertices is the zero vector.
Solution:
Let us assume O be the centre of a regular octagon, as we know that the
centre of a regular octagon bisects all the diagonals passing through it.
So,
…….(i)
…….(ii)
…….(iii)
[Tex]\overrightarrow{OD}=-\overrightarrow{OH} [/Tex] …….(iv)
Now on adding equation (i), (ii), and (iv), we get
Hence proved
Question 9. If P is a point and ABCD is quadrilateral and, show that ABCD is a parallelogram.
Solution:
According to the question
Since,
By using triangle law in ∆APB,
and using triangle law in ∆ DPC,
We get
So, AB is parallel to DC and equal is magnitude.
Hence, ABCD is a parallelogram.
Question 10. Five forces and act at the vertex of a regular hexagon ABCDEF. Prove that the resultant is 6 where o is the centre of hexagon.
Solution:
According to the question,
Prove that
Proof:
As we know that the centre(O) of the hexagon bisects the diagonal
So,
Now,
On adding these equations, we get
⇒
But
So,
Hence proved
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