Vector Algebra Test

Result

Vector Algebra Test - 24

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The vector sum of $$N$$ coplanar forces, having magnitude of $$F$$, when each force is making an angle of $$\displaystyle \frac{2\pi}{N}$$ with that preceding it, is:

A
$$F$$

B
$$\displaystyle \frac{NF}{2}$$

C
$$NF$$

D
$$0$$

Solution

If there are $$N$$ forces of magnitude $$F$$ and each force makes an angle of $$\dfrac { 2 \pi}{N} $$ with the preceding one, then these forces form the sides of a $$N$$ sided regular polygon thus making the resultant of all the forces equal to zero.
Question 2
1 / -0

Five equal forces each of 20N are acting at a point in the same plane. If the angles between them are same, then the resultant of these forces is:

A
0

B
40N

C
20N

D
$$20\sqrt{2}$$

Solution

When we take out components for all the vectors, they all cancel out each other.
So, by assuming it to be a symmetric case (as angles between all forces are same and acting at the same point) we can directly conclude that the resultant is zero.
Question 3
1 / -0

If there are 11 vectors each having a magnitude equal to $$| \vec{p} |$$ and if each side of polygon subtends an angle $$30^{0}$$ at the centre of the polygon. Then the resultant is

A
$$\vec{p}$$

B
0

C
$$\frac{\vec{p}}{2}$$

D
2$$\vec{p}$$

Solution

$$12$$ vectors at $$30^0$$ can make resultant $$\overrightarrow{0}$$ due to circular symmetry.
Vector sum of $$11$$ vector $$+ 12$$ vector $$=0$$
$$R+\overrightarrow{p}$$$$=0$$
$$R=-$$$$\overrightarrow{p}$$
Hence, resultant vector will be opposite direction of $$R$$.
Question 4
1 / -0

$$\mathrm{If}$$ $$\vec{AD},\ \vec{BE},\ \vec{CF}$$ are medians of an equilateral triangle $$\mathrm{A}\mathrm{B}\mathrm{C}$$, then $$\vec{AD}+\vec{BE}+\vec{CF}$$ equals to

A
$$\vec{AB}+\vec{BC}+\vec{CA}$$

B
a zero vector

C
both $$1$$ and $$2$$

D
$$2\vec{AF}+3\vec{BF}$$

Solution

The question wants to ask what is the vector sum of these $$23 $$ median vectors. The answer is correct. Let the vectors of vertices be $$a, b$$ and $$0$$ then the median vectors are $$(a+b)/2, b/2 - a$$ and $$a/2 - b$$
The sum is a zero vector. and $$a+ b-a +(-b) $$ which is vector sum of $$\vec{AB}, \vec{BC}$$ and $$\vec{CA}$$ is also $$0$$.
Question 5
1 / -0

If $$\overrightarrow { a } ,\overrightarrow { b } $$ and $$\overrightarrow { c } $$ are three non-zero vectors such that $$\overrightarrow { a } .\overrightarrow { b } =\overrightarrow { a } .\overrightarrow { c } ,$$ then

A
$$\overrightarrow { b } =\overrightarrow { c } $$

B
$$\overrightarrow { a } \bot \overrightarrow { b } ,\overrightarrow { c } $$

C
$$\overrightarrow { a } \bot \overrightarrow { b } -\overrightarrow { c } $$

D
either $$\overrightarrow { a } \bot (\overrightarrow { b } -\overrightarrow { c }) $$ or $$\overrightarrow { b } =\overrightarrow { c } $$

Solution

     Here If    $$\vec{a} ,\vec{b}$$    and    $$\vec{c}$$
     are   three non-zero vectors   such that

     $$\vec{a} .\vec{b}$$ = $$\vec{a} .\vec{c}$$
     $$\vec{a} .\vec{b}$$ - $$\vec{a} .\vec{c} =0 $$
     $$ \vec{a}.(\vec{b} - \vec{c}) =0 $$
     $$ \vec{a} =0 , (\vec{b} - \vec{c}) =0 $$
     $$ \vec{a} =0$$ or $$\vec{b} = \vec{c} $$
     Therefore from this , we get
     Either $$ \vec{a}\perp (\vec{b} - \vec{c})$$    or   $$\vec{b} = \vec{c}$$
     Hence Option $$(D)$$
Question 6
1 / -0

Let $$\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$$ be a trapezium and let $$\mathrm{P},\mathrm{Q}$$ be the midpoints of the nonparallel sides $$\mathrm{A}\mathrm{D},\ \mathrm{B}\mathrm{C}$$ respectively. Then $$\vec { PQ } $$

A
$$\vec { AB } +\vec { DC }$$

B
$${ \dfrac { 1 }{ 4 } }(\vec { AB } +\vec { BC } )$$

C
$$\dfrac { 1 }{ 2 } (\vec { AB } +\vec { DC } )$$

D
$$\dfrac { 1 }{ 2 } (\vec { AB } +\vec { BC } )$$

Solution

$$AP=PD$$ ($$P$$ is mid point)
$$BQ=QC$$ ($$Q$$ is mid point)

Let $$\vec{a},\vec{b},\vec{c}, \vec{d}$$ are position vector of $$ABCD$$ respectively
position vector of $$P = \dfrac{\vec{a}+\vec{d}}{2}$$
position vector of $$Q = \dfrac{\vec{b}+\vec{c}}{2}$$
$$\vec{PQ}=\dfrac{1}{2}(\vec{b}-\vec{a}-\vec{d}+\vec{c})$$
$$=\dfrac{1}{2}(\vec{AB}-\vec{CD})$$
$$=\dfrac{1}{2}(\vec{AB}+\vec{DC})$$
Question 7
1 / -0

Let $$\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$$ be a parallelogram and let $$\mathrm{L}$$ and $$\mathrm{M}$$ be the midpoints of the sides $${ BC } $$ and $$ { CD }$$ respectively. Then $$\vec { AL } +\vec { AM } =$$

A
$$\vec { AC } $$

B
$${ \dfrac { 2 }{ 3 } }\vec { AC } $$

C
$$\dfrac { 3 }{ 2 } \vec { AC } $$

D
$$2\vec { AC } $$

Solution

$$BL=LC$$
$$CM=DM$$
$$\vec{AB}=\vec{DC}=\vec{a}$$
$$\vec{AD}=\vec{BC}=\vec{b}$$
$$\vec{AL}=\vec{AB}+\vec{BL}$$
$$\vec{AM}=\vec{AD}+\vec{DM}$$
$$\vec{AL}+\vec{AM}=\vec{AB}+\vec{\dfrac{BC}{2}}+\vec{AD}+\vec{\dfrac{DC}{2}}$$
$$=\dfrac{3}{2}\left ( \vec{AB}+\vec{BC} \right )$$
$$=\dfrac{3}{2}\vec{AC}$$
Question 8
1 / -0

$${A}{B}{C}{D}$$ is a quadrilateral, $$E$$ is the point of intersection of the line joining the middle points of the opposite sides. lf $$O$$ is any point, then $$\hat { OA } +\hat { OB } +\hat { OC } +\hat { OD } =$$

A
$$4\hat { OE } $$

B
$$3\hat { OE } $$

C
$$2\hat { OE } $$

D
$$\hat { OE } $$

Solution

Let $$PV$$ (position vector) of $$A,B,C,D$$ be $$\vec{a}, \vec{b}, \vec{c}, \vec{d}$$ respectively.
$$PV$$ of $$P$$ $$=$$ $$ \dfrac{\vec{a}+\vec{b}}{2}$$
$$PV$$ of $$Q$$ $$=$$ $$\dfrac{\vec{c}+\vec{d}}{2}$$
$$PV $$ of $$v$$ $$=$$ $$\dfrac{\vec{a}+\vec{d}}{2}$$
$$PV $$ of $$u$$ $$=$$ $$\dfrac{\vec{b}+\vec{b}}{2}$$
$$\vec{PQ}=\dfrac{1}{2}(\vec{d}-\vec{a}+\vec{c}-\vec{b})$$
$$\vec{uv}=\dfrac{1}{2}(\vec{a}-\vec{b}+\vec{d}-\vec{c})$$
So, $$E=\dfrac{\vec{a}+\vec{b}+\vec+{c}+\vec{d}}{4}$$
Let $$O$$ be origin, so $$\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}=\vec{a}+\vec{b}+\vec{c}+\vec{d}$$
Therefore, $$4\vec{OE}=\vec{a}+\vec{b}+\vec{c}+\vec{d}$$
Question 9
1 / -0

If $$\overline{DA}=\overline{a};\overline{AB}=\overline{b}$$ and $$\overline{CB}=k\overline{a}$$ where $$k>0$$ and $$x, y$$ are the midpoints of $$DB$$ and $$AC$$ respectively such that $$|\overline{a}|=17$$ and $$|\overline{XY}|=4$$, then $$\mathrm{k}=$$

A
$$\displaystyle \dfrac{8}{17}$$

B
$$\displaystyle \dfrac{9}{17}$$

C
$$\displaystyle \dfrac{11}{17}$$

D
$$\displaystyle \dfrac{4}{17}$$

Solution

$$AY=CY$$ ....... $$[y$$ is the midpoint of $$AC]$$
$$DX=BX$$ ....... $$[x$$ is the midpoint of $$DB]$$

Let D be the origin
Position vector(P.V.) of $$A=\vec{a}$$
P.V. of $$B=\vec{a}+\vec{b}$$ ....... $$[\because \overline{AB}=\bar{b}]$$
P.V. of $$C=-(k-1)\vec{a}+\vec{b}$$ ....... $$[\because \overline{CB}=k\bar{a}]$$
P.V. of $$ x=\dfrac{\vec{a}+\vec{b}}{2}$$
P.V. of $$ y=\dfrac{-k\vec{a}+2\vec{a}}{2}+\dfrac{\vec{b}}{2}$$
$$\vec{XY}=\dfrac{\vec{a}-k\vec{a}}{2}$$
$$\left | \vec{XY} \right |=\dfrac{\left | a \right |}{2}\sqrt{(1-k)^{2}}$$
$$\implies \dfrac{8}{17}=1-k$$
$$\implies k=\dfrac{9}{17}$$
Question 10
1 / -0

The incentre of the triangle formed by the points $$ { \hat { i } }+{ \hat { j } }+{ \hat { k } },$$ $$ 4{ \hat { i } }+{ \hat { j } }+{ \hat { k } }$$, and $$4{ \hat { i } }+5{ \hat { j } }+\hat { k }$$ is

A
$$\dfrac { { \hat { i } }+{ \hat { j } }+{ \hat { k } } }{ 3 } $$

B
$${ \hat { i } }+2{ \hat { j } }+3\hat { k } $$

C
$$3{ \hat { i } }+2{ \hat { j } }+\hat { k } $$

D
$${ \hat { i } }+{ \hat { j } }+{ \hat { k } }$$

Solution

Position vector of $$A$$ is $$\hat i+\hat j+\hat k$$, gives coordinates as $$(1,1,1)$$.
Position vector of $$B$$ is $$4\hat i+\hat j+\hat k$$, gives coordinates as $$(4,1,1)$$.
Position vector of $$C$$ is $$4\hat i+5\hat j+\hat k$$, gives coordinates as $$(4,5,1)$$.

$$\vec{AB} = \vec B - \vec A = 3\hat i$$
$$\vec {BC} = \vec C - \vec B = 4\hat j$$
$$\vec {AC} = \vec C - \vec A = 3\hat i+4\hat j$$
Here, we can see that $$\triangle ABC$$ is a right angled triangle with right angle at $$B$$.

$$|\vec{AB}| = \sqrt{3^2} = 3$$
$$|\vec{BC}| = \sqrt{4^2} = 4$$
$$|\vec{AC}| = \sqrt{3^2+4^2} = 5$$

Point of Intersection of angle bisectors can be found using formula:
$$I = \dfrac{|\vec{BC}|\times\vec A + |\vec{CA}|\times\vec B + |\vec{AB}|\times\vec C}{|\vec{AB}| + |\vec{BC}| + |\vec{CA}|}$$

$$I = \dfrac{4\times(\hat i+\hat j+\hat k) + 5\times(4\hat i+\hat j+\hat k) + 3\times(4\hat i+5\hat j+\hat k)}{3+4+5}$$

$$\Rightarrow I = \dfrac{1}{12}(36\hat i + 24\hat j+12\hat j) = 3\hat i+2\hat j+\hat k$$

Hence, incenter will be $$3\hat i+2\hat j +\hat k$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 24

Result

Vector Algebra Test - 24

Score

-

Rank

-

SHARING IS CARING

Question 1

$$F$$

$$\displaystyle \frac{NF}{2}$$

$$NF$$

$$0$$

Solution

Question 2

0

40N

20N

$$20\sqrt{2}$$

Solution

Question 3

$$\vec{p}$$

0

$$\frac{\vec{p}}{2}$$

2$$\vec{p}$$

Solution

Question 4

$$\vec{AB}+\vec{BC}+\vec{CA}$$

a zero vector

both $$1$$ and $$2$$

$$2\vec{AF}+3\vec{BF}$$

Solution

Question 5

$$\overrightarrow { b } =\overrightarrow { c } $$

$$\overrightarrow { a } \bot \overrightarrow { b } ,\overrightarrow { c } $$

$$\overrightarrow { a } \bot \overrightarrow { b } -\overrightarrow { c } $$

either $$\overrightarrow { a } \bot (\overrightarrow { b } -\overrightarrow { c }) $$ or $$\overrightarrow { b } =\overrightarrow { c } $$

Solution

Question 6

$$\vec { AB } +\vec { DC }$$

$${ \dfrac { 1 }{ 4 } }(\vec { AB } +\vec { BC } )$$

$$\dfrac { 1 }{ 2 } (\vec { AB } +\vec { DC } )$$

$$\dfrac { 1 }{ 2 } (\vec { AB } +\vec { BC } )$$

Solution

Question 7

$$\vec { AC } $$

$${ \dfrac { 2 }{ 3 } }\vec { AC } $$

$$\dfrac { 3 }{ 2 } \vec { AC } $$

$$2\vec { AC } $$

Solution

Question 8

$$4\hat { OE } $$

$$3\hat { OE } $$

$$2\hat { OE } $$

$$\hat { OE } $$

Solution

Question 9

$$\displaystyle \dfrac{8}{17}$$

$$\displaystyle \dfrac{9}{17}$$

$$\displaystyle \dfrac{11}{17}$$

$$\displaystyle \dfrac{4}{17}$$

Solution

Question 10

$$\dfrac { { \hat { i } }+{ \hat { j } }+{ \hat { k } } }{ 3 } $$

$${ \hat { i } }+2{ \hat { j } }+3\hat { k } $$

$$3{ \hat { i } }+2{ \hat { j } }+\hat { k } $$

$${ \hat { i } }+{ \hat { j } }+{ \hat { k } }$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.