Accessing OP Malhotra ISC Class 12 Solutions Chapter 21 Vectors Ex 21(a) can be a valuable tool for students seeking extra practice.
S Chand Class 12 ICSE Maths Solutions Chapter 21 Vectors Ex 21(a)
Question 1.
Draw a pair of directed segments \(\overrightarrow{A B}\) and \(\overrightarrow{X Y}\) which are parallel, in the opposite sense, and equal in length. Find a directed segment which represents their sum. What can you say about
(i) its length,
(ii) its direction?
Answer:
Since \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{XY}}\) are equal in length and opposite in sense
So length of this vector be 0 and direction be indeterminate.
Question 2.
Simplify for triangles A B C and P Q R
(i) \(\overrightarrow{\mathbf{A B}}\) + \(\overrightarrow{\mathbf{B A}}\)
(ii) \(\overrightarrow{\mathbf{B C}}\) + \(\overrightarrow{\mathbf{C A}}\) + \(\overrightarrow{\mathbf{A B}}\)
(iii) \(\overrightarrow{\mathbf{P Q}}\) + \(\overrightarrow{\mathbf{R P}}\) + \(\overrightarrow{\mathbf{Q R}}\)
Answer:
(i) \(\overrightarrow{\mathrm{AB}}\) + \(\overrightarrow{\mathrm{BA}}\) = \(\overrightarrow{\mathrm{AB}}\) – \(\overrightarrow{\mathrm{AB}}\) = \(\overrightarrow{0}\)
(ii) By δ law of sum of vectors, we have
Question 3.
In Fig. given below the various line segments are taken to be representatives of vectors. Find from the figure a single representative of each of the following sums :
(i) \(\overrightarrow{\mathbf{A E}}\) + \(\overrightarrow{\mathbf{E C}}\)
(ii) \(\overrightarrow{\mathrm{DB}}\) + \(\overrightarrow{\mathrm{BE}}\)
(iii) \(\overrightarrow{\mathbf{A D}}\) + \(\overrightarrow{\mathrm{DB}}\) + \(\overrightarrow{\mathbf{B C}}\)
(iv) \(\overrightarrow{\mathbf{C B}}\) + \(\overrightarrow{\mathbf{B E}}\) + \(\overrightarrow{\mathbf{E A}}\) + \(\overrightarrow{\mathbf{A D}}\)
Answer:
Question 4.
In Fig. given below, simplify :
(i) \(\overrightarrow{\mathrm{DE}}\) + (\(-\overrightarrow{\mathrm{BE}}\))
(ii) \(\overrightarrow{\mathrm{AC}}\) + (\(-\overrightarrow{\mathrm{BC}}\))
(iii) \(\overrightarrow{\mathrm{CD}}\) + \(\overrightarrow{\mathrm{BA}}\)(\(-\overrightarrow{\mathrm{BD}}\))
Answer:
Question 5.
In Fig. simplify:
(i) \(\overrightarrow{\mathbf{A C}}\) – \(\overrightarrow{\mathbf{A B}}\)
(ii) \(\overrightarrow{\mathbf{B A}}\) – \(\overrightarrow{\mathbf{B C}}\)
(iii) \(\overrightarrow{\mathbf{B A}}\) – \(\overrightarrow{\mathbf{B C}}\)
(iv) \(\overrightarrow{\mathrm{CA}}\) – \(\overrightarrow{\mathrm{CB}}\)
(v) \(\overrightarrow{C B}\) – \(\overrightarrow{C A}\)
Answer:
Question 6.
In Fig. given below, EFGH is a parallelogram. Simplify :
(i) \(\overrightarrow{E F}\) + \(\overrightarrow{E H}\)
(ii) \(\overrightarrow{\mathbf{E F}}\) – \(\overrightarrow{\mathbf{E H}}\)
(iii) \(\overrightarrow{\mathbf{E H}}\) – \(\overrightarrow{\mathbf{E F}}\)
(iv) \(\overrightarrow{F G}\) + \(\overrightarrow{F E}\)
(v) \(\overrightarrow{\mathrm{FE}}\) + \(\overrightarrow{\mathrm{FG}}\)
(vi) \(\overrightarrow{F G}\) – \(\overrightarrow{F E}\)
(vii) \(\overrightarrow{\mathbf{G E}}\) – \(\overrightarrow{\mathbf{G H}}\)
(viii) \(\overrightarrow{\mathrm{HG}}\) – \(\overrightarrow{\mathrm{HE}}\)
Answer:
