1. Scalar Product of Vectors Notes: We can perform addition, subtraction and multiplication of vector quantities. However, division of them is not possible. During multiplication, we may get a scalar result or a vector result depending upon the type of multiplication we choose. …

In this page, you can find the complete solutions of the first exercise of Review of Composition and Resolution of Vectors chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan. In the above-mentioned book, circle is the 13th chapter and has th…

Question: If ABCD is a parallelogram and G is the point of intersection of its diagonals. If O is any point then prove that: $\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD}$$= 4\vec{OG}$ Here's the solution that has been embedded from our official Instagram profile. You can visit it anytime to check the latest solutions that we have …

Question: In the figure, $\vec{OA} = \vec{a}, \vec{OB} = \vec{b} \; and \; \vec{AB} = 3 \vec{AP}$. Find the value of $\vec{OP}$ in terms of $\vec{a}$ and $\vec{b}$. Solution: Given: $\vec{OA} = \vec{a}, \vec{OB} = \vec{b} \; and \; \vec{AB} = 3 \vec{AP}$ To find: value of $\vec{OP}$ In ∆ OAB, Using ∆ law of vector addition, $\vec{A…

Question: In the diagram, $\vec{PQ} = \vec{p}  \; and \;  \vec{QS} = \vec{q}. If \frac{ \vec{QS}}{ \vec{SR}} = \frac{1}{3}$, express $\vec{PR}$ in terms of $\vec{p}$ and $\vec{q}$. Solution: Given: $\vec{PQ} = \vec{p}, \vec{QS} = \vec{q}  \; and \;   \frac{ \vec{QS}}{ \vec{SR}} = \frac{1}{3}$ To find: value of $\vec{PR}$ In ∆ PQS, …

Question: In the figure, $\vec{OA} = \vec{a}, \vec{OB} = \vec{b} \;and\; \vec{AB} = 2\vec{AP}$, find the value of $\vec{OP}$ in terms of $\vec{a}$ and $\vec{b}$. Solution: Given: $\vec{OA} = \vec{a}, \vec{OB} = \vec{b} \; and \; \vec{AB} = 2\vec{AP}$ To find: value of $\vec{b}$ In ∆ OAP, Using ∆ law of vector addition, $\vec{AP} = …

Question: In the given figure, $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$. If $\vec{AC} = 3 \vec{AB}$, find $\vec{OC}$. Solution: Given: $\vec{OA} = \vec{a} \; and \; \vec{OB} = \vec{b}$, $\vec{AC} = 3 \vec{AB}$ To find: value of $\vec{OC}$ In ∆ OAB, Using ∆ law of vector addition, $\vec{AB} = \vec{AO} + \vec{OB}$ $or, \vec{AB} …

Question: In the given figure, ABCD is a parallelogram. If $\vec{OA} = \vec{a}, \vec{OB} = \vec{b}, and \; \vec{OC} = \vec{c}, find \; \vec{OD}$. Solution: To find: value of $\vec{OD}$ Given, $\vec{OA} = \vec{a}, \vec{OB} = \vec{b}, and \; \vec{OC} = \vec{c}$ ABCD is a parallelogram. So, opposite sides are equal and opposite vector…

Question: In the given hexagon ABCDEF, prove that $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{EF} + \vec{FA} = 0$ Solution: To prove: $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{EF} + \vec{FA} = 0$ In a regular hexagon, opposite and parallel vectors in same direction are equal. So, $\vec{AB} = \vec{ED}$ $\vec{FA} = \vec…

Question: In the given figure, prove that: $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} = \vec{AE}$ Solution: To prove: $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} = \vec{AE}$ Construction: Join AC and join CE. Using ∆ law of vector addition in ∆ ABC, $\vec{AC} = \vec{AB} + \vec{BC}$ - (i) Using ∆ law of vector addition in ∆ CDE, …

Question: In the given diagram ABCD is a pentagon, prove that: $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{EA} = 0$ Solution: To prove: $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{EA} = 0$ Using ∆ law of vector addition in ∆ ABC, $\vec{AC} = \vec{AB} + \vec{BC}$ - (i) Using ∆ law of vector addition in ∆ ADE, $\vec{DA} =…

Question: Prove that $\vec{PQ} + \vec{QR} + \vec{RS} + \vec{SP} = 0$ in the given quadrilateral PQRS. Solution: To prove: $\vec{PQ} + \vec{QR} + \vec{RS} + \vec{SP} = 0$ Construction: Join P and R. Proof: Using ∆ law of vector addition, $\vec{PR} = \vec{PQ} + \vec{QR}$ - (i) Also, $\vec{RP} = \vec{RS} + \vec{SP}$ - (ii) Taking …

Question: Prove that $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = 0$ in the given triangle ABC. Solution: To prove: $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = 0$ We know, In ∆ABC, Using triangle law of vector addition, $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$…