OP Malhotra Class 12 Maths Solutions Chapter 21 Vectors Exercise 21 (B)

Question 1. \( \overrightarrow{O A} \) and \( \overrightarrow{O B} \) are vectors \( \vec{a} \) and \( \vec{b} \) respectively and X and Y are points of trisection of A B. Find, in terms of \( \vec{a} \) and \( \vec{b} \).
(i) \( \overrightarrow{\mathrm{OX}} \) and
(ii) \( \overrightarrow{\mathrm{OY}} \)
Answer:

X is a point that divides the line segment AB in the ratio 1:2. The position vector of X, denoted as \( \overrightarrow{\mathrm{OX}} \), can be found using the section formula.
(i) P.V. of X (Position Vector of X) \( = \frac{1 \cdot \vec{b} + 2 \cdot \vec{a}}{1+2} \)
\( \implies \overrightarrow{\mathrm{OX}} = \frac{2\vec{a} + \vec{b}}{3} \)
Also, Y is another point that divides the line segment AB in the ratio 2:1. The position vector of Y, denoted as \( \overrightarrow{\mathrm{OY}} \), can also be found using the section formula. This formula helps us locate points on a line segment when we know the position vectors of its endpoints.
(ii) P.V. of Y \( = \frac{2 \cdot \vec{b} + 1 \cdot \vec{a}}{2+1} \)
\( \implies \overrightarrow{\mathrm{OY}} = \frac{\vec{a} + 2\vec{b}}{3} \)
In simple words: X and Y split the line AB into three equal parts. We use a special formula called the section formula to find where X and Y are located. X is closer to A, and Y is closer to B.

🎯 Exam Tip: Remember the section formula: if a point divides a line segment in ratio m:n, its position vector is \( \frac{n\vec{a} + m\vec{b}}{m+n} \). Pay close attention to which vector is associated with which ratio part.

Question 2. \( \overrightarrow{O A} \) and \( \overrightarrow{O B} \) are vectors \( \vec{a} \) and \( \vec{b} \) respectively and P and Q are points \( \frac{1}{4} \) and \( \frac{3}{4} \) of the way along A B. Find, in terms of \( \vec{a} \) and \( \vec{b} \).
(i) \( \overrightarrow{\mathrm{OP}} \) and (ii) \( \overrightarrow{\mathrm{OQ}} \).
Answer:

(i) Point P is \( \frac{1}{4} \) of the way along AB, which means P divides the line segment AB in the ratio 1:3. Therefore, its position vector \( \overrightarrow{\mathrm{OP}} \) is:
P.V. of P \( = \frac{1 \cdot \vec{b} + 3 \cdot \vec{a}}{1+3} \)
\( \implies \overrightarrow{\mathrm{OP}} = \frac{3\vec{a} + \vec{b}}{4} \)
(ii) Point Q is \( \frac{3}{4} \) of the way along AB, which means Q divides the line segment AB in the ratio 3:1. Its position vector \( \overrightarrow{\mathrm{OQ}} \) is:
P.V. of Q \( = \frac{3 \cdot \vec{b} + 1 \cdot \vec{a}}{3+1} \)
\( \implies \overrightarrow{\mathrm{OQ}} = \frac{\vec{a} + 3\vec{b}}{4} \)
In simple words: P and Q are points on the line AB. P is one-quarter of the way from A to B, and Q is three-quarters of the way from A to B. We use the section formula to find their exact locations, which are expressed using vectors \( \vec{a} \) and \( \vec{b} \).

🎯 Exam Tip: When points are described as "\(\frac{m}{n}\) of the way along AB", it implies the ratio of division is m:(n-m) from A. For example, \( \frac{1}{4} \) means a 1:3 ratio.

Question 3. A B C D is a quadrilateral in which B C is parallel to A D and the ratio of the lengths B C: A D is 4:7. Taking \( \overrightarrow{AB} \) and \( \overrightarrow{AD} \) as representatives of vectors \( \vec{v} \) and \( 7\vec{u} \) respectively, find which vectors are represented by
(i) \( \overrightarrow{\mathrm{BC}} \)
(ii) \( \overrightarrow{\mathbf{A C}} \)
(iii) \( \overrightarrow{\mathrm{BD}} \)
(iv) \( \overrightarrow{\mathrm{DC}} \)
(v) \( \overrightarrow{\mathrm{AE}} \) where E is on BD such that B E = \( \frac{4}{11} \) BD in length;
(vi) \( \overrightarrow{\mathbf{A F}} \) where F is on AC such that AF = \( \frac{7}{11} \) AC.
Answer:

We are given that BC is parallel to AD and the ratio of their lengths is BC:AD = 4:7. We also know \( \overrightarrow{AB} = \vec{v} \) and \( \overrightarrow{AD} = 7\vec{u} \).
(i) Since BC is parallel to AD and their ratio is 4:7, we can write:
\( \overrightarrow{BC} = \frac{4}{7} \overrightarrow{AD} \)
\( \implies \overrightarrow{BC} = \frac{4}{7} (7\vec{u}) \)
\( \implies \overrightarrow{BC} = 4\vec{u} \)
(ii) To find \( \overrightarrow{AC} \), we use the triangle law of vector addition for \( \triangle ABC \):
\( \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} \)
\( \implies \overrightarrow{AC} = \vec{v} + 4\vec{u} \)
(iii) To find \( \overrightarrow{BD} \), we use the triangle law of vector addition for \( \triangle ABD \):
\( \overrightarrow{BA} + \overrightarrow{AD} = \overrightarrow{BD} \)
\( \implies \overrightarrow{BD} = -\overrightarrow{AB} + \overrightarrow{AD} \)
\( \implies \overrightarrow{BD} = -\vec{v} + 7\vec{u} \)
(iv) To find \( \overrightarrow{DC} \), we use the triangle law of vector addition for \( \triangle BDC \):
\( \overrightarrow{DC} + \overrightarrow{CB} = \overrightarrow{DB} \)
\( \implies \overrightarrow{DC} = \overrightarrow{DB} - \overrightarrow{CB} \)
\( \implies \overrightarrow{DC} = - \overrightarrow{BD} + \overrightarrow{BC} \)
\( \implies \overrightarrow{DC} = -(-\vec{v} + 7\vec{u}) + 4\vec{u} \)
\( \implies \overrightarrow{DC} = \vec{v} - 7\vec{u} + 4\vec{u} \)
\( \implies \overrightarrow{DC} = \vec{v} - 3\vec{u} \)
(v) Point E is on BD such that \( BE = \frac{4}{11} BD \). This means E divides BD in the ratio 4:7. Using the section formula from A as the origin:
\( \overrightarrow{AE} = \overrightarrow{AB} + \overrightarrow{BE} \)
\( \implies \overrightarrow{AE} = \overrightarrow{AB} + \frac{4}{11} \overrightarrow{BD} \)
\( \implies \overrightarrow{AE} = \vec{v} + \frac{4}{11} (-\vec{v} + 7\vec{u}) \)
\( \implies \overrightarrow{AE} = \vec{v} - \frac{4}{11}\vec{v} + \frac{28}{11}\vec{u} \)
\( \implies \overrightarrow{AE} = \frac{11\vec{v} - 4\vec{v} + 28\vec{u}}{11} \)
\( \implies \overrightarrow{AE} = \frac{7\vec{v} + 28\vec{u}}{11} \)
\( \implies \overrightarrow{AE} = \frac{7}{11} (\vec{v} + 4\vec{u}) \)
(vi) Point F is on AC such that \( AF = \frac{7}{11} AC \). This means F divides AC in the ratio 7:4. From A as the origin:
\( \overrightarrow{AF} = \frac{7}{11} \overrightarrow{AC} \)
\( \implies \overrightarrow{AF} = \frac{7}{11} (\vec{v} + 4\vec{u}) \)
In simple words: We are given a four-sided shape (quadrilateral) with some vector information. We use the rules of vector addition and ratios to find the vectors representing different sides and internal segments. For instance, if two lines are parallel and we know their length ratio, we can find one vector from the other. Similarly, when a point divides a line in a certain ratio, we can find its vector position.

🎯 Exam Tip: Always draw a clear diagram for vector problems in geometry. Remember that \( \overrightarrow{BA} = -\overrightarrow{AB} \) and the triangle law of vector addition \( \overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PR} \).

Question 4. In fig. given below, B E is median of triangle A B C and G divides B E in the ratio 2 : 1.
(i) If \( \overrightarrow{\mathrm{AB}} \) represents \( \vec{u} \) and \( \overrightarrow{\mathrm{AC}} \) represents \( \vec{v} \), show that \( \overrightarrow{\mathrm{EB}} \) represents \( \vec{u} - \frac{1}{2} \vec{v} \) and \( \overrightarrow{A G} \) represents \( \frac{1}{3}(\vec{u} + \vec{v}) \).
(ii) If C F is a median, and H divides C F in the ratio 2 : 1, show that \( \overrightarrow{\mathrm{AH}} \) represents \( \frac{1}{3}(\vec{u} + \vec{v}) \).
(iii) If AD is a median and K divides AD in the ratio 2 : 1, which vector does \( \overrightarrow{A K} \) represents in terms of \( \vec{u} \) and \( \vec{v} \) ? What can you conclude about G, H, K ? What can you conclude about the medians of a triangle?
Answer:
(i) Since BE is the median of \( \triangle ABC \), E is the midpoint of AC.
Therefore, \( \overrightarrow{AE} = \overrightarrow{EC} = \frac{1}{2} \overrightarrow{AC} = \frac{\vec{v}}{2} \).
Using the triangle law of vectors for \( \triangle AEB \):
\( \overrightarrow{AE} + \overrightarrow{EB} = \overrightarrow{AB} \)
\( \implies \overrightarrow{EB} = \overrightarrow{AB} - \overrightarrow{AE} \)
\( \implies \overrightarrow{EB} = \vec{u} - \frac{\vec{v}}{2} \)

Since G divides BE in the ratio 2:1 (BG:GE = 2:1), taking A as the origin, we use the section formula to find \( \overrightarrow{AG} \):
\( \overrightarrow{AG} = \frac{1 \cdot \overrightarrow{AB} + 2 \cdot \overrightarrow{AE}}{1+2} \)
\( \implies \overrightarrow{AG} = \frac{1 \cdot \vec{u} + 2 \cdot \frac{\vec{v}}{2}}{3} \)
\( \implies \overrightarrow{AG} = \frac{\vec{u} + \vec{v}}{3} \)
(ii) Since CF is the median of \( \triangle ABC \), F is the midpoint of AB.
Therefore, \( \overrightarrow{AF} = \overrightarrow{FB} = \frac{1}{2} \overrightarrow{AB} = \frac{\vec{u}}{2} \).

Taking A as the origin and H divides CF in the ratio 2:1 (CH:HF = 2:1), we use the section formula to find \( \overrightarrow{AH} \):
\( \overrightarrow{AH} = \frac{1 \cdot \overrightarrow{AC} + 2 \cdot \overrightarrow{AF}}{1+2} \)
\( \implies \overrightarrow{AH} = \frac{1 \cdot \vec{v} + 2 \cdot \frac{\vec{u}}{2}}{3} \)
\( \implies \overrightarrow{AH} = \frac{\vec{v} + \vec{u}}{3} \)
(iii) Since AD is the median of \( \triangle ABC \), D is the midpoint of BC.
Therefore, \( \overrightarrow{BD} = \overrightarrow{DC} = \frac{1}{2} \overrightarrow{BC} \).
The position vector of D, taking A as the origin, is:
\( \overrightarrow{AD} = \frac{\overrightarrow{AB} + \overrightarrow{AC}}{2} \)
\( \implies \overrightarrow{AD} = \frac{\vec{u} + \vec{v}}{2} \)

Taking A as the origin and K divides AD in the ratio 2:1 (AK:KD = 2:1), we use the section formula to find \( \overrightarrow{AK} \):
\( \overrightarrow{AK} = \frac{1 \cdot \overrightarrow{AA} + 2 \cdot \overrightarrow{AD}}{1+2} \) (Since A is the origin, \( \overrightarrow{AA} = \vec{0} \))
\( \implies \overrightarrow{AK} = \frac{1 \cdot \vec{0} + 2 \cdot (\frac{\vec{u} + \vec{v}}{2})}{3} \)
\( \implies \overrightarrow{AK} = \frac{\vec{u} + \vec{v}}{3} \)
We can conclude that G, H, and K all represent the same position vector \( \frac{\vec{u} + \vec{v}}{3} \). This means that G, H, and K are all the same point. This point is the centroid of the triangle, which is where all three medians intersect. Therefore, the medians of a triangle are concurrent, meaning they all meet at a single point.
In simple words: For any triangle, if you draw a line from each corner to the middle of the opposite side (these lines are called medians), all three lines will cross at the same point. This special point is called the centroid. We showed this by calculating the vector for this point using three different medians and finding they all give the same result.

🎯 Exam Tip: The centroid of a triangle is the point of concurrency of its medians. Its position vector can always be expressed as \( \frac{\vec{a} + \vec{b} + \vec{c}}{3} \) if the origin is outside the triangle, or if one vertex is the origin, it simplifies to \( \frac{\vec{u} + \vec{v}}{3} \) for the two other vertices' position vectors.

Question 5. Four points A, B, C, D with position vectors \( \vec{a}, \vec{b}, \vec{c}, \vec{d} \) respectively are such that \( 3\vec{a} - \vec{b} + 2\vec{c} - 4\vec{d} = \overrightarrow{0} \). Show that the four points are coplanar. Also, find the position vector of the points of intersection of lines AC and BD.
Answer:

We are given the vector equation: \( 3\vec{a} - \vec{b} + 2\vec{c} - 4\vec{d} = \overrightarrow{0} \)
We can rearrange this equation to group terms:
\( 3\vec{a} + 2\vec{c} = \vec{b} + 4\vec{d} \)
Now, divide both sides by 5 (since 3+2 = 5 and 1+4 = 5):
\( \frac{3\vec{a} + 2\vec{c}}{5} = \frac{\vec{b} + 4\vec{d}}{5} \)
We can write the denominators as sums of ratios:
\( \frac{3\vec{a} + 2\vec{c}}{3+2} = \frac{1\vec{b} + 4\vec{d}}{1+4} \)
The left side represents the position vector of a point that divides AC in the ratio 2:3. Let's call this point P. So, \( \overrightarrow{OP} = \frac{3\vec{a} + 2\vec{c}}{3+2} \).
The right side represents the position vector of a point that divides BD in the ratio 4:1. This is also \( \overrightarrow{OP} = \frac{1\vec{b} + 4\vec{d}}{1+4} \).
Since both expressions give the same position vector P, it means that point P lies on both line segment AC and line segment BD. Therefore, P is the point of intersection of AC and BD.
Since P is a common point on both diagonals, the lines AC and BD intersect, and all four points A, B, C, and D must lie on the same plane (coplanar).
The position vector of the point of intersection of lines AC and BD is \( \frac{3\vec{a} + 2\vec{c}}{5} \) or \( \frac{\vec{b} + 4\vec{d}}{5} \).
In simple words: We started with a vector equation involving four points. By rearranging it, we found that a specific point P can be defined in two ways: it divides line AC in one ratio and also divides line BD in another ratio. Since the point is the same, it means the lines AC and BD cross each other at P. If the lines formed by these four points cross, then all four points must lie on the same flat surface, making them coplanar.

🎯 Exam Tip: To prove that points are coplanar from a vector equation like \( l\vec{a} + m\vec{b} + n\vec{c} + p\vec{d} = \vec{0} \), rearrange it into the form \( l\vec{a} + n\vec{c} = -m\vec{b} - p\vec{d} \). If the sums of coefficients are equal (i.e., \( l+n = -m-p \)), then the ratio representation will show a common point of intersection, proving coplanarity.