If D is the midpoint of the side BC of a triangle – Class 11 Mathematics | ConceptFirst

Exercise 8.1 · Q9

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Concept understanding — Algebra of Vectors

Addition — two equivalent pictures.

Triangle law: if $\vec a=\vec{A_1B_1}$ and $\vec b=\vec{B_1B_2}$ (the tail of $\vec b$ placed at the tip of $\vec a$ ), then $\vec a+\vec b$ is the third side $\vec{A_1B_2}$ , taken from the start of $\vec a$ to the end of $\vec b$ . In words: if two vectors are represented, in magnitude and direction, by two sides of a triangle taken in order, their sum is the third side taken in the reverse order.
Parallelogram law: if $\vec a=\vec{OA}$ and $\vec b=\vec{OB}$ share the initial point $O$ , complete the parallelogram $OACB$ ; the diagonal $\vec{OC}$ through $O$ is $\vec a+\vec b$ .

Both laws describe the same sum — the triangle law is just the parallelogram law applied to half of the parallelogram.

Key results proved from the triangle law:

If $\vec a,\vec b,\vec c$ are the three sides of a triangle taken in order (tip to tail, returning to the start), $\vec a+\vec b+\vec c=\vec 0$ .
Vector addition is associative: $(\vec a+\vec b)+\vec c=\vec a+(\vec b+\vec c)$ .
$\vec a+\vec 0=\vec 0+\vec a=\vec a$ for every $\vec a$ .
$\vec a+(-\vec a)=\vec 0$ , where $-\vec a$ (the reverse of $\vec a$ ) has the same magnitude as $\vec a$ but the opposite direction; if $\vec a=\vec{AB}$ then $-\vec a=\vec{BA}$ .
Vector addition is commutative: $\vec a+\vec b=\vec b+\vec a$ (proved by the parallelogram, since both diagonals lead to the same point $C$ ).
Polygon law: for any chain of vectors placed tip to tail, $\vec{OA}+\vec{AB}+\vec{BC}+\vec{CD}+\vec{DE}=\vec{OE}$ — the sum is the single vector from the very first tail to the very last tip. …

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Step 1. Take $A$ as origin, so $\vec{AB}=\vec b,\ \vec{AC}=\vec c$ .

Step 2. $D$ is the midpoint of $BC$ , so its position vector (w.r.t. $A$ ) is $\vec{AD}=\dfrac{\vec b+\vec c}{2}=\dfrac{\vec{AB}+\vec{AC}}2$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{DA}+\overrightarrow{CD}$
(a) $\vec{0}$
(b) $\overrightarrow{AD}$
(c) $-\overrightarrow{AD}$
(d) $\overrightarrow{CA}$
›Reveal solutionSolution

Using the triangle law repeatedly, $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ and $\overrightarrow{CD}+\overrightarrow{DA}=\overrightarrow{CA}$ , and these two cancel to give $\vec{0}$ .

$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{DA}+\overrightarrow{CD}$

Group the first two: $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ (triangle law: consecutive vectors add head-to-tail).

…

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CBSE 2025Set ANNUAL1 markMCQ
$\vec{a} + 2\vec{b}$
(a) $6$
(b) $3$
(c) $\dfrac{1}{6}$
(d) $\dfrac{1}{3}$
›Reveal solutionSolution

Two vectors expressed in the same basis $\vec a,\vec b$ are parallel exactly when their coefficients are proportional.

If $\vec a+2\vec b$ is parallel to $3\vec a+m\vec b$ , then $3\vec a+m\vec b=k(\vec a+2\vec b)$ for some scalar $k$ . …

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CBSE 2023Set ANNUAL1 markMCQ
$\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{DA} + \overrightarrow{CD}$
(a) $\vec{0}$
(b) $\overrightarrow{AD}$
(c) $-\overrightarrow{AD}$
(d) $\overrightarrow{CA}$
›Reveal solutionSolution

Rearranging the four vectors as $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}$ traces a closed path $A\to B\to C\to D\to A$ , which always sums to zero.

Vector addition does not depend on the order in which the vectors are added:

$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{DA}+\overrightarrow{CD} = \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}$

…

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CBSE 2022Set ANNUAL1 markMCQ
$\vec{a}$
(a) $\vec{a} + \vec{b}$
(b) $\vec{a} - \vec{b}$
(c) $\dfrac{\vec{a} + \vec{b}}{2}$
(d) $\vec{b} - \vec{a}$
›Reveal solutionSolution

With $\overrightarrow{AB}=\vec a$ and $\overrightarrow{AD}=\vec b$ as adjacent sides, the other diagonal is $\overrightarrow{BD}=\vec b-\vec a$ .

In parallelogram $ABCD$ , take $A$ as the origin. Then $\overrightarrow{AB}=\vec a$ and $\overrightarrow{AD}=\vec b$ are the adjacent sides, and the diagonal $\overrightarrow{AC}=\vec a+\vec b$ is given, consistent with $C = A+\vec a+\vec b$ (opposite vertex).

…

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CBSE 2020Set ANNUAL1 markMCQ
$\hat{i} + \hat{j} - \hat{k}$
(a) $\dfrac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{5}}$
(b) $\dfrac{2\hat{i} + \hat{j}}{\sqrt{5}}$
(c) $\dfrac{2\hat{i} - \hat{j} + \hat{k}}{\sqrt{5}}$
(d) $\dfrac{2\hat{i} - \hat{j}}{\sqrt{5}}$
›Reveal solutionSolution

The resultant vector is $2\hat i-\hat j$ ; dividing by its magnitude $\sqrt5$ gives the unit vector.

The resultant (sum) of the two vectors is found by adding corresponding components:

$(\hat i+\hat j-\hat k)+(\hat i-2\hat j+\hat k)=(1+1)\hat i+(1-2)\hat j+(-1+1)\hat k=2\hat i-\hat j+0\hat k=2\hat i-\hat j.$

Its magnitude is $\sqrt{2^2+(-1)^2}=\sqrt{4+1}=\sqrt5$ .

…

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CBSE 2019Set ANNUAL1 markMCQ
$\hat{i}+\hat{j}-\hat{k}$
(a) $\dfrac{2\hat{i}-\hat{j}+\hat{k}}{\sqrt{5}}$
(b) $\dfrac{2\hat{i}-\hat{j}}{\sqrt{5}}$
(c) $\dfrac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{5}}$
(d) $\dfrac{2\hat{i}+\hat{j}}{\sqrt{5}}$
›Reveal solutionSolution

The resultant of $\hat{i}+\hat{j}-\hat{k}$ and $\hat{i}-2\hat{j}+\hat{k}$ is $2\hat{i}-\hat{j}$ (the $\hat{k}$ components cancel); dividing by its magnitude $\sqrt{5}$ gives the unit vector.

Resultant $= (\hat{i}+\hat{j}-\hat{k}) + (\hat{i}-2\hat{j}+\hat{k})$ .

Add componentwise: $\hat{i}$ -component: $1+1=2$ ; $\hat{j}$ -component: $1-2=-1$ ; $\hat{k}$ -component: $-1+1=0$ .

Resultant $= 2\hat{i}-\hat{j}$ .

…

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CBSE 2018Set ANNUAL1 markMCQ
$\vec{a}$
(a) $\dfrac{1}{|\vec{a}|}$
(b) $|\vec{a}|$
(c) $\pm\dfrac{1}{|\vec{a}|}$
(d) $1$
›Reveal solutionSolution

$|k\vec a|=|k||\vec a|=1$ gives $|k|=1/|\vec a|$ , so $k$ can be either $+1/|\vec a|$ or $-1/|\vec a|$ .

For a scalar $k$ and vector $\vec a$ , $|k\vec a| = |k|\cdot|\vec a|$ (magnitude of a scaled vector is the absolute value of the scalar times the vector's magnitude).

Given $|k\vec a|=1$ : $|k|\cdot|\vec a| = 1 \Rightarrow |k| = \dfrac{1}{|\vec a|} \Rightarrow k = \pm\dfrac{1}{|\vec a|}$

…

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