MCQ 11 Mark
The position vectors of the points A, B, C are $2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\ 3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$ respectively. These points,
- ✓
Form an isosceles triangle.
- B
- C
- D
AnswerCorrect option: A. Form an isosceles triangle.
Given: Position vectors of A, B, C are $2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\ 3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$. Then,
$\overrightarrow{\text{AB}}=\big(3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)-\big(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
$=\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
$\overrightarrow{\text{BC}}=\big(\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}\big)-\big(3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)$
$=-2\hat{\text{i}}+6\hat{\text{j}}-4\hat{\text{k}}$
$\overrightarrow{\text{CA}}=\big(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)-\big(\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}\big)$
$=\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
Now, $\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{1^2+(-3)^2+2^2}$
$=\sqrt{1+9+4}$
$=\sqrt{14}$
$\Big|\overrightarrow{\text{CA}}\Big|=\sqrt{1^2+(-3)^2+2^2}$
$=\sqrt{1+9+4}$
$=\sqrt{14}$
$\Big|\overrightarrow{\text{BC}}\Big|=\sqrt{(-2)^2+6^2+(-4)^2}$
$=\sqrt{4+36+16}$
$=\sqrt{56}$
$\therefore\Big|\overrightarrow{\text{AB}}\Big|=\Big|\overrightarrow{\text{CA}}\Big|$
Hence, the triangle is isosceles as two of its sides are equal.
View full question & answer→MCQ 21 Mark
If G is the intersection of diagonals of a parallelogram ABCD and O is any point, then $\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}=$
- A
$2\overrightarrow{\text{OG}}$
- ✓
$4\overrightarrow{\text{OG}}$
- C
$5\overrightarrow{\text{OG}}$
- D
$3\overrightarrow{\text{OG}}$
AnswerCorrect option: B. $4\overrightarrow{\text{OG}}$
Let us consider the point O as origin.
G is the mid-point of AC.
$\therefore\ \overrightarrow{\text{OG}}=\frac{\overrightarrow{\text{OA}}+\overrightarrow{\text{OC}}}2$
$2\overrightarrow{\text{OG}}=\overrightarrow{\text{OA}}+\overrightarrow{\text{OC}}\ \dots(1)$
Also, G is the mid-point BD
$\therefore\ \overrightarrow{\text{OG}}=\frac{\overrightarrow{\text{OB}}+\overrightarrow{\text{OD}}}2$
$2\overrightarrow{\text{OG}}=\overrightarrow{\text{OB}}+\overrightarrow{\text{OD}}\ \dots(2)$
On adding (1) and (2) we get,
$2\overrightarrow{\text{OG}}+2\overrightarrow{\text{OG}}=\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}$
$4\overrightarrow{\text{OG}}=\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}$
$\therefore\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}=4\overrightarrow{\text{OG}}$ View full question & answer→MCQ 31 Mark
ABCD is a parallelogram with AC and BD as diagonals. Then, $\overrightarrow{\text{AC}}-\overrightarrow{\text{BD}}=$
- A
$4\overrightarrow{\text{AB}}$
- B
$3\overrightarrow{\text{AB}}$
- ✓
$2\overrightarrow{\text{AB}}$
- D
$\overrightarrow{\text{AB}}$
AnswerCorrect option: C. $2\overrightarrow{\text{AB}}$
Given: ABCD, a parallelogram with diagonals AC and BD. Then,
$\overrightarrow{\text{AC}}=\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}$
$\overrightarrow{\text{AD}}=\overrightarrow{\text{AB}}+\overrightarrow{\text{BD}}$
$\Rightarrow\ \overrightarrow{\text{BD}}=\overrightarrow{\text{AD}}-\overrightarrow{\text{AB}}$
$\therefore\overrightarrow{\text{AC}}-\overrightarrow{\text{BD}}=\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{AD}}+\overrightarrow{\text{AB}}=2\overrightarrow{\text{AB}}$ $\Big[\because\ \overrightarrow{\text{AD}}=\overrightarrow{\text{BC}}\Big]$
View full question & answer→MCQ 41 Mark
In a regular hexagon ABCDEF, $\overrightarrow{\text{AB}}=\vec{\text{a}},\ \overrightarrow{\text{BC}}=\vec{\text{b}}$ and $\overrightarrow{\text{CD}}=\vec{\text{c}}$. Then, $\overrightarrow{\text{AE}}=$
- A
$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$
- B
$2\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$
- ✓
$\vec{\text{b}}+\vec{\text{c}}$
- D
$\vec{\text{a}}+2\vec{\text{b}}+2\vec{\text{c}}$
AnswerCorrect option: C. $\vec{\text{b}}+\vec{\text{c}}$
Given a regular hexagon ABCDEF, $\overrightarrow{\text{AB}}=\vec{\text{a}},\ \overrightarrow{\text{BC}}=\vec{\text{b}}$ and $\overrightarrow{\text{CD}}=\vec{\text{c}}$. Then,
In $\triangle{\text{ABC}}$, we have
$\overrightarrow{\text{AC}}=\vec{\text{a}}+\vec{\text{b}}$
In $\triangle{\text{ACD}}$, we have
$\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}=\overrightarrow{\text{AD}}$
$\Rightarrow\overrightarrow{\text{AD}}=\overrightarrow{\text{AC}}+\vec{\text{c}}$
$\Rightarrow\overrightarrow{\text{AD}}=\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$
Again, in $\triangle{\text{ADE}}$, we have
$\overrightarrow{\text{AE}}=\overrightarrow{\text{AD}}+\overrightarrow{\text{DE}}$
$\Rightarrow\overrightarrow{\text{AE}}=\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}-\vec{\text{a}}$
$\Rightarrow\overrightarrow{\text{AE}}=\vec{\text{b}}+\vec{\text{c}}$
Hence option (c).
View full question & answer→MCQ 51 Mark
If $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are two collinear vectors, then which of the follwoing are incorrect?
- A
$\vec{\text{b}}=\lambda\vec{\text{a}}$ for some scalar $\lambda$
- B
$\vec{\text{a}}=\pm\vec{\text{b}}$
- C
The respective components of $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are proportional.
- ✓
Both the vectors $\vec{\text{a}}\text{ and }\vec{\text{b}}$ have the same direction but different magnitudes.
AnswerCorrect option: D. Both the vectors $\vec{\text{a}}\text{ and }\vec{\text{b}}$ have the same direction but different magnitudes.
If $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are collinear vectors, then they are parallel. Therefore, we have $\vec{\text{b}}=\lambda\vec{\text{a}}$, for some scalar $\lambda$.
If $\lambda=\pm1\Rightarrow\vec{\text{a}}=\pm\vec{\text{b}}$
If $\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ and $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$. Then,
$\vec{\text{b}}=\lambda\vec{\text{a}}$
$\Rightarrow\ \text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}=\lambda\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)$
$\Rightarrow\ \text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}=(\lambda\text{a}_1)\hat{\text{i}}+(\lambda\text{a}_2)\hat{\text{j}}+(\lambda\text{a}_3)\hat{\text{k}}$
$\Rightarrow\ \text{b}_1=\lambda\text{a}_1,\ \text{b}_2=\lambda\text{a}_2,\ \text{b}_3=\lambda\text{a}_3$
$\Rightarrow\ \frac{\text{b}_1}{\text{a}_1}=\frac{\text{b}_2}{\text{a}_2}=\frac{\text{b}_3}{\text{a}_3}=\lambda$
Thus, the respective components of $\vec{\text{a}}\text{ and }\vec{\text{b}}$ can have different directions. Hence, the statement given in (d) is incorrect.
View full question & answer→MCQ 61 Mark
Let $G$ be the centroid of $\triangle{\text{ABC}}$. if $\overrightarrow{\text{AB}}=\vec{\text{a}},\overrightarrow{\text{AC}}=\vec{\text{b}}$, then the bisector $\overrightarrow{\text{AG}}$, in terms of $\vec{\text{a}}$ and $\vec{\text{b}}$ is,
- A
$\frac{2}3\big(\vec{\text{a}}+\vec{\text{b}}\big)$
- B
$\frac{1}6\big(\vec{\text{a}}+\vec{\text{b}}\big)$
- ✓
$\frac{1}3\big(\vec{\text{a}}+\vec{\text{b}}\big)$
- D
$\frac{1}2\big(\vec{\text{a}}+\vec{\text{b}}\big)$
AnswerCorrect option: C. $\frac{1}3\big(\vec{\text{a}}+\vec{\text{b}}\big)$
Taking $A$ as origin. Then, position vector of $A, B$ and $C$ are $\vec0,\vec{\text{a}}$ and $\vec{\text{b}}$ respectively.
Then, Centroid $G$ has position vector $\frac{\vec0+\vec{\text{a}}+\vec{\text{b}}}3=\frac{\vec{\text{a}}+\vec{\text{b}}}3$
Therefore, $\text{AG}=\frac{\vec{\text{a}}+\vec{\text{b}}}3-\vec0=\frac{\vec{\text{a}}+\vec{\text{b}}}3$
View full question & answer→MCQ 71 Mark
If three points A, B and C have position vectors $\hat{\text{i}}+\text{x}\hat{\text{j}}+3\hat{\text{k}},\ 3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}$ and $\text{y}\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}$ respectively are collinear, then (x, y) =
AnswerGiven position vector of A, B and C are $\hat{\text{i}}+\text{x}\hat{\text{j}}+3\hat{\text{k}},\ 3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}$ and $\text{y}\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}$. Then,
$\overrightarrow{\text{AB}}=3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}-\hat{\text{i}}-\text{x}\hat{\text{j}}-3\hat{\text{k}}$
$=2\hat{\text{i}}+(4-\text{x})\hat{\text{j}}+4\hat{\text{k}}$
$\overrightarrow{\text{BC}}=\text{y}\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}-3\hat{\text{i}}-4\hat{\text{j}}-7\hat{\text{k}}$
$=(\text{y}-3)\hat{\text{i}}-6\hat{\text{j}}-12\hat{\text{k}}$
Since, the given vectors are collinear.
$\therefore\ \overrightarrow{\text{AB}}=\lambda\overrightarrow{\text{BC}}$
$\Rightarrow\ 2\hat{\text{i}}+(4-\text{x})\hat{\text{j}}+4\hat{\text{k}}=\lambda(\text{y}-3)\hat{\text{i}}-6\lambda\hat{\text{j}}-12\lambda\hat{\text{k}}$
$\Rightarrow\ 2=\lambda(\text{y}-3),\ (4-\text{x})=-6\lambda,\ 4=-12\lambda$
$\Rightarrow\ 2=\lambda(\text{y}-3),\ (4-\text{x})=-6\lambda,\ \lambda=-\frac{1}3$
$\Rightarrow\ 2=-\frac{1}3(\text{y}-3),\ (4-\text{x})=-6\times\Big(-\frac{1}3\Big)$
$\Rightarrow\ -6=\text{y}-3,\ 4-\text{x}=2$
$\Rightarrow\ \text{y}=-3,\ \text{x}=2$
View full question & answer→MCQ 81 Mark
If $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are three non $-$ zero vectors, no two of which are collinear and the vector $\vec{\text{a}}+\vec{\text{b}}$ is collinear with $\vec{\text{c}}$, $\vec{\text{b}}+\vec{\text{c}}$ is collinear with $\vec{\text{a}}$, then $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=$
- A
$\vec{\text{a}}$
- B
$\vec{\text{b}}$
- C
$\vec{\text{c}}$
- ✓
View full question & answer→MCQ 91 Mark
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ and $\vec{\text{d}}$ are the position vector of points A, B, C, D such that no three of them are collinear and $\vec{\text{a}}+\vec{\text{c}}=\vec{\text{b}}+\vec{\text{d}}$, then ABCD is a,
AnswerGiven:
$\vec{\text{a}}+\vec{\text{c}}=\vec{\text{b}}+\vec{\text{d}}$
$\Rightarrow\vec{\text{c}}-\vec{\text{d}}=\vec{\text{b}}-\vec{\text{a}}$
$\Rightarrow\overrightarrow{\text{AB}}=\overrightarrow{\text{DC}}$
And $\vec{\text{a}}+\vec{\text{c}}=\vec{\text{b}}+\vec{\text{d}}$
$\Rightarrow\vec{\text{c}}-\vec{\text{b}}=\vec{\text{d}}-\vec{\text{a}}$
$\Rightarrow\overrightarrow{\text{AD}}=\overrightarrow{\text{BC}}$
Also, since $\vec{\text{a}}+\vec{\text{c}}=\vec{\text{b}}+\vec{\text{d}}$
$\Rightarrow\frac{1}2\big(\vec{\text{a}}+\vec{\text{c}}\big)=\frac{1}2\big(\vec{\text{b}}+\vec{\text{d}}\big)$
So, position vector of mid-point of BD = position vector of mid-point of AC.
Hence diagonals bisect each other.
The given ABCD is a parallelogram.
View full question & answer→MCQ 101 Mark
Forces $3\overrightarrow{\text{OA}},\ 5\overrightarrow{\text{OB}}$ act along OA and OB. If their resultant passes through C on AB, then,
- A
- B
C divides AB in the ratio 2 : 1
- ✓
- D
AnswerDraw ON, the perpendicular to the line AB
Let $\vec{\text{i}}$ be the unit vector along ON
The resultant force $\vec{\text{R}}=3\overrightarrow{\text{OA}}+5\overrightarrow{\text{OB}}\ \dots(1)$
The angles between $\vec{\text{i}}$ and the forces $\vec{\text{R}},\ 3\overrightarrow{\text{OA}},\ 5\overrightarrow{\text{OB}}$ are $\angle\text{CON},\ \angle\text{AON},\ \angle\text{BON}$ respectively.
$\vec{\text{R}}.\vec{\text{i}}=3\overrightarrow{\text{OA}}.\vec{\text{i}}+5\overrightarrow{\text{OB}}.\vec{\text{i}}$
$\Rightarrow\text{R}.1.\cos\angle\text{CON}\\=3\overrightarrow{\text{OA}}.1.\cos\angle\text{AON}+5\overrightarrow{\text{OB}}.1.\cos\angle{\text{BON}}$
$\text{R}.\frac{\text{ON}}{\text{OC}}=3\text{OA}\times\frac{\text{ON}}{\text{OA}}+5\text{OB}\frac{\text{ON}}{\text{OB}}$
$\frac{\text{R}}{\text{OC}}=(3+5)$
$\text{R}=8\overrightarrow{\text{OC}}$
We know that,
$\overrightarrow{\text{OA}}=\overrightarrow{\text{OC}}+\overrightarrow{\text{CA}}$
$\Rightarrow3\overrightarrow{\text{OA}}=3\overrightarrow{\text{OC}}+3\overrightarrow{\text{CA}}\ \dots(\text{i})$
$\overrightarrow{\text{OB}}=\overrightarrow{\text{OC}}+\overrightarrow{\text{CB}}$
$\Rightarrow5\overrightarrow{\text{OB}}=5\overrightarrow{\text{OC}}+5\overrightarrow{\text{CB}}\ \dots(\text{ii})$
On adding (i) and (ii) we get,
$3\overrightarrow{\text{OA}}+5\overrightarrow{\text{OB}}=8\overrightarrow{\text{OC}}+3\overrightarrow{\text{CA}}+5\overrightarrow{\text{CB}}$
$\vec{\text{R}}=8\overrightarrow{\text{OC}}+3\overrightarrow{\text{CA}}+5\overrightarrow{\text{CB}}$
$8\overrightarrow{\text{OC}}=8\overrightarrow{\text{OC}}+3\overrightarrow{\text{CA}}+5\overrightarrow{\text{CB}}$
$\Big|3\overrightarrow{\text{AC}}\Big|=\Big|5\overrightarrow{\text{CB}}\Big|$
$\Rightarrow3\overrightarrow{\text{AC}}=5\overrightarrow{\text{CB}}$ View full question & answer→MCQ 111 Mark
If points $\text{A}(60\hat{\text{i}}+3\hat{\text{j}}),\ \text{B}(40\hat{\text{i}}-8\hat{\text{j}})$ and $\text{C}(\text{a}\hat{\text{i}}+52\hat{\text{j}})$ are collinear, then a is equal to,
AnswerGiven: Three points $\text{A}(60\hat{\text{i}}+3\hat{\text{j}}),\ \text{B}(40\hat{\text{i}}-8\hat{\text{j}})$ and $\text{C}(\text{a}\hat{\text{i}}+52\hat{\text{j}})$ are collinear. Then,
$\overrightarrow{\text{AB}}=\lambda\overrightarrow{\text{BC}}$
We have,
$\overrightarrow{\text{AB}}=\big(40\hat{\text{i}}-8\hat{\text{j}}\big)-\big(60\hat{\text{i}}+3\hat{\text{j}}\big)$
$=-20\hat{\text{i}}-11\hat{\text{j}}$
$\overrightarrow{\text{BC}}=\big(\text{a}\hat{\text{i}}-52\hat{\text{j}}\big)-\big(40\hat{\text{i}}-8\hat{\text{j}}\big)$
$=(\text{a}-40)\hat{\text{i}}-44\hat{\text{j}}$
Therefore,
$\overrightarrow{\text{AB}}=\lambda\overrightarrow{\text{BC}}$
$\Rightarrow-20\hat{\text{i}}-11\hat{\text{j}}=\lambda(\text{a}-40)\hat{\text{i}}-\lambda44\hat{\text{j}}$
$\Rightarrow\lambda(\text{a}-40)=-20,\ -44\lambda=-11\Rightarrow\lambda=\frac{1}4$
$\Rightarrow\text{a}-40=-80$
$\Rightarrow\text{a}=-40$
View full question & answer→MCQ 121 Mark
If in a $\triangle\text{ABC}$, $\text{A}=(0,0),\ \text{B}=(3,3\sqrt3),\ \text{C}=(-3\sqrt3,3)$, then the vecctor of magnitude $2\sqrt2$ units directed along AO, where O is the circumcenter of $\triangle\text{ABC}$ is,
- ✓
$(1-\sqrt3)\hat{\text{i}}+(1+\sqrt3)\hat{\text{j}}$
- B
$(1+\sqrt3)\hat{\text{i}}+(1-\sqrt3)\hat{\text{j}}$
- C
$(1+\sqrt3)\hat{\text{i}}+(\sqrt3-1)\hat{\text{j}}$
- D
AnswerCorrect option: A. $(1-\sqrt3)\hat{\text{i}}+(1+\sqrt3)\hat{\text{j}}$
$\Big|\overrightarrow{\text{AO}}\Big|=2\sqrt2$
$\Big|\overrightarrow{\text{AO}}\Big|=\Big|\overrightarrow{\text{BO}}\Big|=\Big|\overrightarrow{\text{CO}}\Big|=2\sqrt2=\text{R}$
Let the position vector of O be $\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}$
$\Big|\overrightarrow{\text{AO}}\Big|=\sqrt{\text{x}^2+\text{y}^2}$
$\therefore\ \text{x}^2+\text{y}^2=8\ \dots(1)$
Also, $\Big|\overrightarrow{\text{BO}}\Big|=\Big|\overrightarrow{\text{CO}}\Big|$
$\sqrt{(\text{x}-3)^2+(\text{y}-3\sqrt3)^2}\\=\sqrt{(\text{x}+3\sqrt3)^2+(\text{y}-3)^2}$
$\text{x}^2-6\text{x}+9+\text{y}^2-6\sqrt3\text{y}+27\\=\text{x}^2+6\sqrt3\text{x}+27+\text{y}^2-6\text{y}+9$
$\text{y}(6-6\sqrt3)=\text{x}(6\sqrt3+6)$
$\text{y}=\frac{\text{x}(1+\sqrt3)}{(1-\sqrt3)}\ \dots(2)$
Substituting y from (2) in (1) we get,
$(1-\sqrt3)^2\text{x}^2+(1+\sqrt3)^2\text{x}^2=8(1-\sqrt3)^2$
$\text{x}^2\times8=8(1-\sqrt3)^2$
$\text{x}=1-\sqrt3$
$\text{y}=1+\sqrt3$
$\therefore$ The position vector of O is $(1-\sqrt3)\hat{\text{i}}+(1+\sqrt3)\hat{\text{j}}$
$\overrightarrow{\text{AO}}=(1-\sqrt3)\hat{\text{i}}+(1+\sqrt3)\hat{\text{j}}$
View full question & answer→MCQ 131 Mark
If $\ce{OACB}$ is a parallelogram with $\overrightarrow{\text{OC}}=\vec{\text{a}}$ and $\overrightarrow{\text{AB}}=\vec{\text{b}}$, then $\overrightarrow{\text{OA}}=$
- A
$\big(\vec{\text{a}}+\vec{\text{b}}\big)$
- B
$\big(\vec{\text{a}}-\vec{\text{b}}\big)$
- C
$\frac{1}2\big(\vec{\text{b}}-\vec{\text{a}}\big)$
- ✓
$\frac{1}2\big(\vec{\text{a}}-\vec{\text{b}}\big)$
AnswerCorrect option: D. $\frac{1}2\big(\vec{\text{a}}-\vec{\text{b}}\big)$
Given a parallelogram $\ce{OABC}$ such that $\overrightarrow{\text{OC}}=\vec{\text{a}}$ and $\overrightarrow{\text{AB}}=\vec{\text{b}}$.
Then, $\overrightarrow{\text{OB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{OC}}$
$\Rightarrow \overrightarrow{\text{OB}}=\overrightarrow{\text{OC}}-\overrightarrow{\text{BC}}$
$\Rightarrow \overrightarrow{\text{OB}}=\overrightarrow{\text{OC}}-\overrightarrow{\text{OA}}$ $\Big[\because\overrightarrow{\text{BC}}=\overrightarrow{\text{OA}}\Big]$
$\Rightarrow \overrightarrow{\text{OB}}=\vec{\text{a}}-\overrightarrow{\text{OA}}\ \dots(1)$
Therefore, $\overrightarrow{\text{OA}}+\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}$
$\Rightarrow \overrightarrow{\text{OA}}+\vec{\text{b}}=\vec{\text{a}}-\overrightarrow{\text{OA}} [$Using $(1)]$
$\Rightarrow 2\overrightarrow{\text{OA}}=\vec{\text{a}}-\vec{\text{b}}$
$\Rightarrow \overrightarrow{\text{OA}}=\frac{1}2\big(\vec{\text{a}}-\vec{\text{b}}\big)$
View full question & answer→MCQ 141 Mark
In figure, which of the following is not true?
- A
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\vec0$
- B
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{AC}}=\vec0$
- ✓
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{CA}}=\vec0$
- D
$\overrightarrow{\text{AB}}-\overrightarrow{\text{CB}}+\overrightarrow{\text{CA}}=\vec0$
AnswerCorrect option: C. $\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{CA}}=\vec0$
We have, LHS = $\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{AC}}=\overrightarrow{\text{AC}}-\overrightarrow{\text{CA}}$ $\Big[\because\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}\Big]$
$=-\overrightarrow{\text{CA}}-\overrightarrow{\text{CA}}$
$=-2\overrightarrow{\text{CA}}$
So, $\text{LHS}\neq\text{RHS}$
Hence, It is not true.
View full question & answer→MCQ 151 Mark
If $\vec{\text{a}},\ \vec{\text{b}}$ are the vectors forming consecutive sides of a regular hexagon ABCDEF, then the vector representing side CD is,
- A
$\vec{\text{a}}+\vec{\text{b}}$
- B
$\vec{\text{a}}-\vec{\text{b}}$
- ✓
$\vec{\text{b}}-\vec{\text{a}}$
- D
$-\big(\vec{\text{a}}+\vec{\text{b}}\big)$
AnswerCorrect option: C. $\vec{\text{b}}-\vec{\text{a}}$
Let ABCDEF be a regular hexagon such that $\overrightarrow{\text{AB}}=\vec{\text{a}}$ and $\overrightarrow{\text{BC}}=\vec{\text{b}}$. We know, AD is parallel to BC such that AD = 2BC.
$\therefore\ \overrightarrow{\text{AD}}=2\overrightarrow{\text{BC}}=2\vec{\text{b}}$
In $\triangle{\text{ABC}}$, we have
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}$
$\Rightarrow\ \vec{\text{a}}+\vec{\text{b}}=\overrightarrow{\text{AC}}$
In $\triangle{\text{ACD}}$, we have
$\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}=\overrightarrow{\text{AD}}$
$\Rightarrow\overrightarrow{\text{CD}}=\overrightarrow{\text{AD}}-\overrightarrow{\text{AC}}$
$\Rightarrow\overrightarrow{\text{CD}}=2\vec{\text{b}}-\big(\vec{\text{a}}+\vec{\text{b}}\big)$
$\Rightarrow\overrightarrow{\text{CD}}=\vec{\text{b}}-\vec{\text{a}}$
View full question & answer→MCQ 161 Mark
The vector equation of the plane passing through $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}},$ is $\vec{\text{r}}=\alpha\vec{\text{a}}+\beta\vec{\text{b}}+\gamma\vec{\text{c}}$, provided that,
AnswerCorrect option: B. $\alpha+\beta+\gamma=1$
Given: A plane passing through $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$.
⇒ Lines $\vec{\text{a}}-\vec{\text{b}}$ and $\vec{\text{c}}-\vec{\text{a}}$ lie on the plane.
The parmetric equation of the plane can be written as:
$\vec{\text{r}}=\vec{\text{a}}+\lambda_1\big(\vec{\text{a}}-\vec{\text{b}}\big)+\lambda_2\big(\vec{\text{c}}-\vec{\text{a}}\big)$
$\vec{\text{r}}=\vec{\text{a}}(1+\lambda_1+\lambda_2)-\lambda_1\vec{\text{b}}+\lambda_2\vec{\text{c}}$
Given that $\vec{\text{r}}=\alpha\vec{\text{a}}+\beta\vec{\text{b}}+\gamma\vec{\text{c}}$
$\therefore\alpha+\beta+\gamma=1+\lambda_1-\lambda_2-\lambda_1+\lambda_2$
$\alpha+\beta+\gamma=1$
View full question & answer→MCQ 171 Mark
The vector $\cos\alpha\cos\beta\hat{\text{i}}+\cos\alpha\sin\beta\hat{\text{j}}+\sin\alpha\hat{\text{k}}$ is a,
AnswerGiven: The vector $\cos\alpha\cos\beta\hat{\text{i}}+\cos\alpha\sin\beta\hat{\text{j}}+\sin\alpha\hat{\text{k}}$. Then,
$\big|\cos\alpha\cos\beta\hat{\text{i}}+\cos\alpha\sin\beta\hat{\text{j}}+\sin\alpha\hat{\text{k}}\big|$
$=\sqrt{\cos^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta+\sin^2\alpha}$
$=\sqrt{\cos^2\alpha+\sin^2\alpha}=1$
View full question & answer→MCQ 181 Mark
If O and O' are circumcenter and orthocenter of $\triangle{\text{ABC}}$ , then $\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}$ equals,
- A
$2\overrightarrow{\text{OO}'}$
- ✓
$\overrightarrow{\text{OO}'}$
- C
$\overrightarrow{\text{O}'\text{O}}$
- D
$2\overrightarrow{\text{O}'\text{O}}$
AnswerCorrect option: B. $\overrightarrow{\text{OO}'}$
Given: O be the circumcentre an O' be the orthocenter of $\triangle{\text{ABC}}$. Let G be the centroid of the triangle.
We know that O, G and H are collinear and by geometry $\overrightarrow{\text{O}'\text{G}}=2\overrightarrow{\text{OG}}$. This yields, $\overrightarrow{\text{O}'\text{O}}=\overrightarrow{\text{O}'\text{G}}+\overrightarrow{\text{GO}}=2\overrightarrow{\text{GO}}+\overrightarrow{\text{GO}}=3\overrightarrow{\text{GO}}$
In other words $\overrightarrow{\text{OO}'}=3\overrightarrow{\text{GO}}$
Since, $\overrightarrow{\text{OG}}=\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3$
$\therefore\overrightarrow{\text{OO}'}=3\times\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3=\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$
$=\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}$
View full question & answer→