OP Malhotra Class 12 Maths Solutions Chapter 21 Vectors Exercise 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 parallel, have equal length, and are in opposite directions, their sum is a zero vector, which means they cancel each other out.
(i) The length of the resultant vector is 0 units.
(ii) The direction of the resultant vector is indeterminate (not defined), because a zero vector has no specific direction.In simple words: When you add two vectors that are exactly the same length but point in opposite ways, they cancel each other out. So, the total length becomes zero, and because there's no length, there's no direction to talk about.

🎯 Exam Tip: When vectors are equal in magnitude but opposite in direction, their sum is always a zero vector, often denoted as \( \overrightarrow{0} \).

Question 2. Simplify for triangles A B C and P Q R
(i) \( \overrightarrow{\mathbf{A B}} + \overrightarrow{\mathbf{B A}} \)
(ii) \( \overrightarrow{\mathbf{BC}} + \overrightarrow{\mathbf{C A}} + \overrightarrow{\mathbf{AB}} \)
(iii) \( \overrightarrow{\mathbf{P Q}} + \overrightarrow{\mathbf{R P}} + \overrightarrow{\mathbf{Q R}} \)
Answer:
(i) We know that \( \overrightarrow{\mathrm{BA}} = - \overrightarrow{\mathrm{AB}} \). So,
\( \overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BA}} = \overrightarrow{\mathrm{AB}} - \overrightarrow{\mathrm{AB}} = \overrightarrow{0} \)
(ii) Using the triangle law of vector addition for \( \triangle ABC \):
\( \overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} = \overrightarrow{\mathrm{AC}} \)
Now, substitute this into the given expression:
\( (\overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}}) + \overrightarrow{\mathrm{CA}} = \overrightarrow{\mathrm{AC}} + \overrightarrow{\mathrm{CA}} \)
We know that \( \overrightarrow{\mathrm{CA}} = - \overrightarrow{\mathrm{AC}} \). So,
\( \overrightarrow{\mathrm{AC}} + (-\overrightarrow{\mathrm{AC}}) = \overrightarrow{0} \)
(iii) Rearranging the terms for clarity:
\( \overrightarrow{\mathrm{PQ}} + \overrightarrow{\mathrm{QR}} + \overrightarrow{\mathrm{RP}} \)
Using the triangle law of vector addition for \( \triangle PQR \):
\( \overrightarrow{\mathrm{PQ}} + \overrightarrow{\mathrm{QR}} = \overrightarrow{\mathrm{PR}} \)
Substitute this back into the expression:
\( \overrightarrow{\mathrm{PR}} + \overrightarrow{\mathrm{RP}} \)
We know that \( \overrightarrow{\mathrm{RP}} = - \overrightarrow{\mathrm{PR}} \). So,
\( \overrightarrow{\mathrm{PR}} + (-\overrightarrow{\mathrm{PR}}) = \overrightarrow{0} \)
In simple words: When you add vectors that form a closed shape, like going all the way around a triangle, the total sum is zero. This happens because you end up back where you started.

🎯 Exam Tip: Remember that \( \overrightarrow{\mathrm{AB}} = - \overrightarrow{\mathrm{BA}} \). Also, the triangle law states that if three vectors form the sides of a triangle taken in order, their sum is the zero vector.

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) \( \overrightarrow{\mathbf{C B}} + \overrightarrow{\mathbf{B E}} + \overrightarrow{\mathbf{E A}} + \overrightarrow{\mathbf{AD}} \)
Answer:
(i) Using the triangle law of vector addition for \( \triangle AEC \), if we go from A to E, and then E to C, the total displacement is directly from A to C.
\( \overrightarrow{\mathrm{AE}} + \overrightarrow{\mathrm{EC}} = \overrightarrow{\mathrm{AC}} \)
(ii) Using the triangle law of vector addition for \( \triangle DEB \), if we go from D to B, and then B to E, the total displacement is directly from D to E.
\( \overrightarrow{\mathrm{DB}} + \overrightarrow{\mathrm{BE}} = \overrightarrow{\mathrm{DE}} \)
(iii) First, consider \( \overrightarrow{\mathrm{AD}} + \overrightarrow{\mathrm{DB}} \). Using the triangle law for \( \triangle ADB \), this sum is \( \overrightarrow{\mathrm{AB}} \).
So, the expression becomes \( \overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} \).
Now, using the triangle law for \( \triangle ABC \), this sum is \( \overrightarrow{\mathrm{AC}} \).
Therefore, \( \overrightarrow{\mathrm{AD}} + \overrightarrow{\mathrm{DB}} + \overrightarrow{\mathrm{BC}} = \overrightarrow{\mathrm{AC}} \)
(iv) Let's rearrange the terms to apply the triangle law more easily:
\( (\overrightarrow{\mathrm{CB}} + \overrightarrow{\mathrm{BE}}) + (\overrightarrow{\mathrm{EA}} + \overrightarrow{\mathrm{AD}}) \)
Using the triangle law, \( \overrightarrow{\mathrm{CB}} + \overrightarrow{\mathrm{BE}} = \overrightarrow{\mathrm{CE}} \).
And \( \overrightarrow{\mathrm{EA}} + \overrightarrow{\mathrm{AD}} = \overrightarrow{\mathrm{ED}} \).
So, the expression simplifies to \( \overrightarrow{\mathrm{CE}} + \overrightarrow{\mathrm{ED}} \).
Using the triangle law again for \( \triangle CED \), this sum is \( \overrightarrow{\mathrm{CD}} \).
Therefore, \( \overrightarrow{\mathrm{CB}} + \overrightarrow{\mathrm{BE}} + \overrightarrow{\mathrm{EA}} + \overrightarrow{\mathrm{AD}} = \overrightarrow{\mathrm{CD}} \)
In simple words: When you add vectors, imagine walking along paths. If you go from point A to E, then E to C, the total journey is like going directly from A to C. This idea helps simplify chains of vectors into a single vector.

🎯 Exam Tip: The head-to-tail rule (triangle law) is key: if the endpoint of one vector is the starting point of the next, their sum is a vector from the first starting point to the last endpoint.

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:
(i) First, we change the subtraction to addition by reversing the vector: \( -\overrightarrow{\mathrm{BE}} = \overrightarrow{\mathrm{EB}} \).
So, the expression becomes \( \overrightarrow{\mathrm{DE}} + \overrightarrow{\mathrm{EB}} \).
Now, using the triangle law of vector addition for \( \triangle DEB \), if we go from D to E, and then E to B, the result is the vector from D to B.
Therefore, \( \overrightarrow{\mathrm{DE}} + \overrightarrow{\mathrm{EB}} = \overrightarrow{\mathrm{DB}} \)
(ii) First, we change the subtraction to addition: \( -\overrightarrow{\mathrm{BC}} = \overrightarrow{\mathrm{CB}} \).
So, the expression becomes \( \overrightarrow{\mathrm{AC}} + \overrightarrow{\mathrm{CB}} \).
Now, using the triangle law of vector addition for \( \triangle ACB \), if we go from A to C, and then C to B, the result is the vector from A to B.
Therefore, \( \overrightarrow{\mathrm{AC}} + \overrightarrow{\mathrm{CB}} = \overrightarrow{\mathrm{AB}} \)
(iii) First, change the subtraction to addition: \( -\overrightarrow{\mathrm{BD}} = \overrightarrow{\mathrm{DB}} \).
So, the expression becomes \( \overrightarrow{\mathrm{CD}} + \overrightarrow{\mathrm{BA}} + \overrightarrow{\mathrm{DB}} \).
Rearrange the terms to apply the triangle law:
\( \overrightarrow{\mathrm{CD}} + \overrightarrow{\mathrm{DB}} + \overrightarrow{\mathrm{BA}} \)
Using the triangle law for \( \triangle CDB \), \( \overrightarrow{\mathrm{CD}} + \overrightarrow{\mathrm{DB}} = \overrightarrow{\mathrm{CB}} \).
So, the expression simplifies to \( \overrightarrow{\mathrm{CB}} + \overrightarrow{\mathrm{BA}} \).
Using the triangle law for \( \triangle CBA \), this sum is \( \overrightarrow{\mathrm{CA}} \).
Therefore, \( \overrightarrow{\mathrm{CD}} + \overrightarrow{\mathrm{BA}} + (-\overrightarrow{\mathrm{BD}}) = \overrightarrow{\mathrm{CA}} \)
In simple words: When you subtract a vector, it's like adding the same vector but in the opposite direction. Then, you can use the triangle law to find a simpler single vector that represents the total journey.

🎯 Exam Tip: Remember that \( -\overrightarrow{\mathrm{PQ}} = \overrightarrow{\mathrm{QP}} \). This sign change allows you to reverse the direction of a vector, which is often useful for applying the triangle law for simplification.

Question 5. In Fig. simplify:
(i) \( \overrightarrow{\mathbf{A C}} - \overrightarrow{\mathbf{A B}} \)
(ii) \( \overrightarrow{\mathbf{B A}} - \overrightarrow{\mathbf{B C}} \)
(iii) \( \overrightarrow{\mathbf{B C}} - \overrightarrow{\mathbf{B A}} \)
(iv) \( \overrightarrow{\mathrm{CA}} - \overrightarrow{\mathrm{CB}} \)
(v) \( \overrightarrow{\mathrm{C B}} - \overrightarrow{\mathrm{C A}} \)
Answer:
(i) We rewrite the subtraction as addition: \( \overrightarrow{\mathrm{AC}} - \overrightarrow{\mathrm{AB}} = \overrightarrow{\mathrm{AC}} + (-\overrightarrow{\mathrm{AB}}) = \overrightarrow{\mathrm{AC}} + \overrightarrow{\mathrm{BA}} \).
Using the commutative property, this is \( \overrightarrow{\mathrm{BA}} + \overrightarrow{\mathrm{AC}} \).
By the triangle law of vector addition for \( \triangle ABC \), \( \overrightarrow{\mathrm{BA}} + \overrightarrow{\mathrm{AC}} = \overrightarrow{\mathrm{BC}} \).
(ii) We rewrite the subtraction as addition: \( \overrightarrow{\mathrm{BA}} - \overrightarrow{\mathrm{BC}} = \overrightarrow{\mathrm{BA}} + (-\overrightarrow{\mathrm{BC}}) = \overrightarrow{\mathrm{BA}} + \overrightarrow{\mathrm{CB}} \).
Using the commutative property, this is \( \overrightarrow{\mathrm{CB}} + \overrightarrow{\mathrm{BA}} \).
By the triangle law of vector addition for \( \triangle CBA \), \( \overrightarrow{\mathrm{CB}} + \overrightarrow{\mathrm{BA}} = \overrightarrow{\mathrm{CA}} \).
(iii) We rewrite the subtraction as addition: \( \overrightarrow{\mathrm{BC}} - \overrightarrow{\mathrm{BA}} = \overrightarrow{\mathrm{BC}} + (-\overrightarrow{\mathrm{BA}}) = \overrightarrow{\mathrm{BC}} + \overrightarrow{\mathrm{AB}} \).
Using the commutative property, this is \( \overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} \).
By the triangle law of vector addition for \( \triangle ABC \), \( \overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} = \overrightarrow{\mathrm{AC}} \).
(iv) We rewrite the subtraction as addition: \( \overrightarrow{\mathrm{CA}} - \overrightarrow{\mathrm{CB}} = \overrightarrow{\mathrm{CA}} + (-\overrightarrow{\mathrm{CB}}) = \overrightarrow{\mathrm{CA}} + \overrightarrow{\mathrm{BC}} \).
Using the commutative property, this is \( \overrightarrow{\mathrm{BC}} + \overrightarrow{\mathrm{CA}} \).
By the triangle law of vector addition for \( \triangle BCA \), \( \overrightarrow{\mathrm{BC}} + \overrightarrow{\mathrm{CA}} = \overrightarrow{\mathrm{BA}} \).
(v) We rewrite the subtraction as addition: \( \overrightarrow{\mathrm{CB}} - \overrightarrow{\mathrm{CA}} = \overrightarrow{\mathrm{CB}} + (-\overrightarrow{\mathrm{CA}}) = \overrightarrow{\mathrm{CB}} + \overrightarrow{\mathrm{AC}} \).
Using the commutative property, this is \( \overrightarrow{\mathrm{AC}} + \overrightarrow{\mathrm{CB}} \).
By the triangle law of vector addition for \( \triangle ACB \), \( \overrightarrow{\mathrm{AC}} + \overrightarrow{\mathrm{CB}} = \overrightarrow{\mathrm{AB}} \).
In simple words: When vectors start from the same point, like \( \overrightarrow{BA} \) and \( \overrightarrow{BC} \), subtracting them gives you a vector that connects their endpoints, in this case, \( \overrightarrow{CA} \). Remember that \( \overrightarrow{X} - \overrightarrow{Y} \) is the same as \( \overrightarrow{X} + (-\overrightarrow{Y}) \).

🎯 Exam Tip: A useful identity for subtraction is \( \overrightarrow{\mathrm{XY}} - \overrightarrow{\mathrm{XZ}} = \overrightarrow{\mathrm{ZY}} \). This means if two vectors share a common starting point, their difference is the vector connecting their endpoints, reversing the order for the subtracted vector.

Question 6. In Fig. given below, EFGH is a parallelogram. Simplify :
(i) \( \overrightarrow{\mathrm{E F}} + \overrightarrow{\mathrm{E H}} \)
(ii) \( \overrightarrow{\mathbf{E F}} - \overrightarrow{\mathbf{E H}} \)
(iii) \( \overrightarrow{\mathbf{E H}} - \overrightarrow{\mathbf{E F}} \)
(iv) \( \overrightarrow{\mathrm{F G}} + \overrightarrow{\mathrm{F E}} \)
(v) \( \overrightarrow{\mathrm{FG}} - \overrightarrow{\mathrm{FE}} \)
(vi) \( \overrightarrow{\mathrm{F G}} - \overrightarrow{\mathrm{F E}} \)
(vii) \( \overrightarrow{\mathrm{G E}} - \overrightarrow{\mathrm{G H}} \)
(viii) \( \overrightarrow{\mathrm{HG}} - \overrightarrow{\mathrm{HE}} \)
Answer:
(i) In parallelogram EFGH, the sum of two adjacent vectors starting from the same point gives the diagonal. So, using the parallelogram law of vector addition:
\( \overrightarrow{\mathrm{E F}} + \overrightarrow{\mathrm{E H}} = \overrightarrow{\mathrm{EG}} \)
(ii) We rewrite the subtraction as addition: \( \overrightarrow{\mathbf{E F}} - \overrightarrow{\mathbf{E H}} = \overrightarrow{\mathbf{E F}} + (-\overrightarrow{\mathbf{E H}}) = \overrightarrow{\mathbf{E F}} + \overrightarrow{\mathbf{H E}} \).
Rearranging terms: \( \overrightarrow{\mathbf{H E}} + \overrightarrow{\mathbf{E F}} \).
By the triangle law of vector addition for \( \triangle HEF \), this sum is \( \overrightarrow{\mathbf{H F}} \).
(iii) We rewrite the subtraction as addition: \( \overrightarrow{\mathbf{E H}} - \overrightarrow{\mathbf{E F}} = \overrightarrow{\mathbf{E H}} + (-\overrightarrow{\mathbf{E F}}) = \overrightarrow{\mathbf{E H}} + \overrightarrow{\mathbf{F E}} \).
Rearranging terms: \( \overrightarrow{\mathbf{F E}} + \overrightarrow{\mathbf{E H}} \).
By the triangle law of vector addition for \( \triangle FEH \), this sum is \( \overrightarrow{\mathbf{F H}} \).
(iv) In parallelogram EFGH, the sum of two adjacent vectors starting from the same point gives the diagonal. So, using the parallelogram law of vector addition:
\( \overrightarrow{\mathrm{F G}} + \overrightarrow{\mathrm{F E}} = \overrightarrow{\mathrm{FH}} \)
(v) We rewrite the subtraction as addition: \( \overrightarrow{\mathrm{FG}} - \overrightarrow{\mathrm{FE}} = \overrightarrow{\mathrm{FG}} + (-\overrightarrow{\mathrm{FE}}) = \overrightarrow{\mathrm{FG}} + \overrightarrow{\mathrm{EF}} \).
Rearranging terms: \( \overrightarrow{\mathrm{EF}} + \overrightarrow{\mathrm{FG}} \).
By the triangle law of vector addition for \( \triangle EFG \), this sum is \( \overrightarrow{\mathrm{EG}} \).
(vi) This is the same expression as (v): \( \overrightarrow{\mathrm{F G}} - \overrightarrow{\mathrm{F E}} \). The result is the same as in (v).
We rewrite the subtraction as addition: \( \overrightarrow{\mathrm{FG}} - \overrightarrow{\mathrm{FE}} = \overrightarrow{\mathrm{FG}} + (-\overrightarrow{\mathrm{FE}}) = \overrightarrow{\mathrm{FG}} + \overrightarrow{\mathrm{EF}} \).
Rearranging terms: \( \overrightarrow{\mathrm{EF}} + \overrightarrow{\mathrm{FG}} \).
By the triangle law of vector addition for \( \triangle EFG \), this sum is \( \overrightarrow{\mathrm{EG}} \).
(vii) We rewrite the subtraction as addition: \( \overrightarrow{\mathrm{GE}} - \overrightarrow{\mathrm{GH}} = \overrightarrow{\mathrm{GE}} + (-\overrightarrow{\mathrm{GH}}) = \overrightarrow{\mathrm{GE}} + \overrightarrow{\mathrm{HG}} \).
Rearranging terms: \( \overrightarrow{\mathrm{HG}} + \overrightarrow{\mathrm{GE}} \).
By the triangle law of vector addition for \( \triangle HGE \), this sum is \( \overrightarrow{\mathrm{HE}} \).
(viii) We rewrite the subtraction as addition: \( \overrightarrow{\mathrm{HG}} - \overrightarrow{\mathrm{HE}} = \overrightarrow{\mathrm{HG}} + (-\overrightarrow{\mathrm{HE}}) = \overrightarrow{\mathrm{HG}} + \overrightarrow{\mathrm{EH}} \).
Rearranging terms: \( \overrightarrow{\mathrm{EH}} + \overrightarrow{\mathrm{HG}} \).
By the triangle law of vector addition for \( \triangle EHG \), this sum is \( \overrightarrow{\mathrm{EG}} \).
In simple words: In a parallelogram, opposite sides are equal and parallel. This means \( \overrightarrow{EF} = \overrightarrow{HG} \) and \( \overrightarrow{EH} = \overrightarrow{FG} \). When simplifying vector expressions, you can change subtraction to addition by reversing the vector's direction. Then, use the triangle law to combine vectors that go from head to tail.

🎯 Exam Tip: When simplifying vectors in a parallelogram, always look for opportunities to use the triangle law (head-to-tail addition) or the parallelogram law (sum of two adjacent vectors from a common point gives the diagonal originating from that point).