Vector Algebra Test

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Vector Algebra Test - 31

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Question 1
1 / -0

If $$G$$ is the centroid of a $$\Delta ABC$$, then $$\vec{GA} + \vec{GB} + \vec{GC}$$ is equal to

A
$$\vec{0}$$

B
$$3\vec{GA}$$

C
$$3\vec{GB}$$

D
$$3\vec{GC}$$

Solution

We have $$\vec{GB}+\vec{GC} = (1+1) \vec{GD} = 2 \vec{GD}$$, where $$D$$ is the midpoint of $$BC$$. Therefore,
$$\vec{GA}+ \vec{GB} + \vec{GC} = \vec{GA} + 2\vec{GD}$$
$$= \vec{GA} - \vec{GA} =0$$
$$(\because G$$ divides $$AC$$ in the ration $$2 : 1, \therefore 2 \vec{GD} = - \vec{GA})$$
Question 2
1 / -0

In triangle ABC, which of the following is not true.

A
$$\overrightarrow{AB} + \overrightarrow{BC}+\overrightarrow{CA} = \overrightarrow{0}$$

B
$$\overrightarrow{AB} + \overrightarrow{BC}- \overrightarrow{AC} = \overrightarrow{0}$$

C
$$\overrightarrow{AB} + \overrightarrow{BC}- \overrightarrow{CA} = \overrightarrow{0}$$

D
$$\overrightarrow{AB} - \overrightarrow{CB} + \overrightarrow{CA} = \overrightarrow{0}$$

Solution

$${\textbf{Hint: Use Triangle Law of vector addition}}$$
$${\textbf{Step 1: Find the resultant of two vetors}}$$
$$ \Rightarrow \overrightarrow {AB}  + \overrightarrow {BC}  = \overrightarrow {AC} $$
$${\textbf{Step 2: Solve the vector sum and Compare with the options}}$$
$$ \overrightarrow {AB}  + \overrightarrow {BC}  - \overrightarrow {AC}  = \overrightarrow 0 $$
$${\Rightarrow}\overrightarrow {AB}  + \overrightarrow {BC}  + \overrightarrow {CA}  = \overrightarrow 0 $$
$${\Rightarrow}\overrightarrow {AB} - \overrightarrow {CB}  + \overrightarrow {CA}  = \overrightarrow 0 $$
$${\text{But }}$$$$\overrightarrow {AB}  + \overrightarrow {BC}  - \overrightarrow {CA}  \ne \overrightarrow 0 $$
$${\textbf{Final Answer: Option (C) is incorrect}}$$
Question 3
1 / -0

Six vectors, $$\vec a$$ to $$\vec f$$ , all of magnitude 1 and directions indicated in the figure ( Consider all of them to be originating at origin ). Which of the following statement is true?

A
$$\vec b+\vec e=\vec f$$

B
$$\vec b+\vec c=\vec f$$

C
$$\vec d+\vec c=\vec f$$

D
$$\vec d+\vec e=\vec f$$

Solution

We have, assuming unit magnitude for all vectors:
$$ \overrightarrow{a}= \overrightarrow{i} $$
$$ \overrightarrow{b}= \overrightarrow{-j} $$
$$ \overrightarrow{c}= \overrightarrow{i} $$
$$ \overrightarrow{d}= \overrightarrow{j} $$
$$ \overrightarrow{e}= \overrightarrow{-i} $$
$$ \overrightarrow{f}= \overrightarrow{-i} + \overrightarrow{j} $$
$$\therefore \overrightarrow{d} + \overrightarrow{e} = \overrightarrow{j} + \overrightarrow{-i} $$
Question 4
1 / -0

In triangle $$ABC$$, $$\angle A = 30^o$$, $$H$$ is the orthocentre and $$D$$ is the midpoint of $$BC$$. Segment $$HD$$ is produced to $$T$$ such that $$HD = DT$$. The length $$AT$$ is equal to

A
$$2BC$$

B
$$3BC$$

C
$$\displaystyle \dfrac{4}{3}BC$$

D
None of these

Solution

Let us assume that the circumcenter of $$\triangle ABC$$ is $$O$$ and is situated at the origin of the coordinate plane. Hence, the vector notation for all the vertices is
$$\overrightarrow{OA}=\overrightarrow{a}$$
$$\overrightarrow{OB}=\overrightarrow{b}$$
$$\overrightarrow{OC}=\overrightarrow{c}$$
Also, as $$O$$ is the circumcenter of $$\triangle ABC$$,
$$\overrightarrow{a}=\overrightarrow{b}=\overrightarrow{c}= R$$ (circum-radius)
Since $$D$$ is mid-point of $$BC$$, $$\overrightarrow{d}=\dfrac{\overrightarrow{b}+\overrightarrow{c}}{2}$$
$$\therefore \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{a}+2\overrightarrow{d}$$
Also, $$2\overrightarrow{OD}=\overrightarrow{AH}$$ as $$H$$ is the orthocentre and it divides height in 2:1 ratio.
$$\therefore \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{a}+2\overrightarrow{d}=\overrightarrow{OA}+\overrightarrow{AH} = \overrightarrow{OH}=\overrightarrow{h}\ \dots\ (1)$$
According to given information, $$HD=DT$$
$$\implies \dfrac{\overrightarrow{b}+\overrightarrow{c}}{2}=\dfrac{\overrightarrow{h}+\overrightarrow{t}}{2}$$
$$\implies \dfrac{\overrightarrow{b}+\overrightarrow{c}}{2}=\dfrac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{t}}{2}$$
$$\implies \overrightarrow{a}=-\overrightarrow{t}$$
$$\implies \overrightarrow{AT}=2\overrightarrow{a}$$
$$\implies |\overrightarrow{AT}|=2|\overrightarrow{a}|=2R$$
Also, from property of side-length of triangle,
$$BC=2RsinA=R$$ ($$\because\angle A = 30^{\circ}$$)
$$\therefore AT=2BC$$
Question 5
1 / -0

$$'I'$$ is the incentre of triangle of $$ABC$$ whose corresponding sides are $$a, b, c,$$ respectively, $$a\vec{IB} + b\vec{IB} + c\vec{IC}$$ is always equal to

A
$$\vec{0}$$

B
$$(a+b+c) \vec{BC}$$

C
$$(\vec{a} + \vec{b} + \vec{c}) \vec{AC}$$

D
$$(a + b + c)\vec{AB}$$

Solution

Let the incentre be at the origin and be
$$A(\vec{p}), B(\vec{q}) $$ and $$C(\vec{r})$$., then

$$\vec{IA} = \vec{p}, \vec{IB} = \vec{q}$$ and $$\vec{IC} = \vec{r}$$.

Incentre $$I$$ is $$\displaystyle \dfrac{a\vec{p} + b\vec{q}+c\vec{r}}{a + b + c}$$, where $$p = BC, q = AC$$ and $$r = AB$$

Incentre is at the origin. Therefore,
$$\displaystyle \dfrac{a \vec{p} + b\vec{q} + c\vec{r}}{a + b + c} = \vec{0} $$, or
$$a\vec{p} + b\vec{q} + c\vec{r} = 0$$
$$\Rightarrow a\vec{IA} + b\vec{IB} + c\vec{IC} = \vec{0}$$
Question 6
1 / -0

Let $$\vec{a}, \vec{b}$$ and $$\vec{c}$$ be unit vectors such that $$\vec{a} + \vec{b} - \vec{c} = 0$$. If the area of triangle formed by vectors $$\vec{a}$$ and $$\vec{b}$$ is $$A$$, then what is the value of $$4A^2$$?

A
$$3$$

B
$$9$$

C
$$ \dfrac { 3 }{ 4 } $$

D
$$ \dfrac { 9 }{ 4 } $$

Solution

Given, $$\vec{a} + \vec{b} = \vec{c}$$

Now vector $$\vec{c}$$ is along the diagonal of the parallelogram which has adjacent side vectors $$\vec{a}$$ and $$\vec{b}$$.

Since $$\vec{c}$$ is also a unit vector, triangle formed by vectors $$\vec{a}$$ and $$\vec{b}$$ is an equilateral triangle, then
Area of triangle $$A= \displaystyle \dfrac{\sqrt{3}}{4}$$

$$4A^2=4 \left(\dfrac{\sqrt{3}}{4} \right)^2=\dfrac{3}{4}$$
Question 7
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In a trapezium, vector $$\vec{BC} = \alpha \vec{AD}$$. Also, $$\vec p = \vec{AC} + \vec{BD}$$ is collinear with $$\vec{AD}$$ and $$\vec p = \mu \vec{AD}$$, then which of the following is true?

A
$$\mu = \alpha + 2$$

B
$$\mu + \alpha =1$$

C
$$\alpha = \mu + 1$$

D
$$\mu = \alpha +1$$

Solution

Let $$A$$ be the origin.
$$\vec BC = \vec c - \vec b = \alpha \vec d$$ and $$\vec p = \vec{AC} + \vec{BD} = \mu AD$$
Hence, $$\vec p = \vec c + \vec d - \vec b = \mu \vec d$$
Using $$\vec c - \vec b = \alpha \vec d$$
$$\Rightarrow \alpha + 1 = \mu$$
Question 8
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$$P(\vec{p})$$ and $$Q(\vec{q})$$ are the position vectors of two fixed points and $$R(\vec{r})$$ is the position vector of a variable point. If $$R$$ moves such that $$(\vec{r} - \vec{p}) \times (\vec{r} - \vec{q}) = \vec{0}$$, then the locus of $$R$$ is

A
A plane containing the origin $$O$$ and parallel to two non-collinear vectors $$\vec{OP}$$ and $$\vec{OQ}$$

B
The surface of a sphere described on $$PQ$$ as its diameter

C
A line passing through points $$P$$ and $$Q$$

D
A set of lines parallel to line $$PQ$$

Solution

$$(\vec{r}-\vec{p})\times(\vec{r}-\vec{q})=0$$

$$\vec{r}\times(\vec{r}-\vec{q})-\vec{p}\times(\vec{r}-\vec{q})=0$$

$$\vec{r}\times\vec{r}-\vec{r}\times\vec{q}-\vec{p}\times\vec{r}+\vec{p}\times\vec{q}=0$$

$$\vec{p}\times\vec{q}-\vec{r}\times\vec{q}-\vec{p}\times\vec{r}=0$$

$$\vec{p}\times\vec{q}-\vec{r}\times\vec{q}+\vec{r}\times\vec{p}=0$$

$$\vec{p}\times\vec{q}+\vec{r}\times(\vec{p}-\vec{q})=0$$

$$\vec{p}\times\vec{q}=\vec{r}\times(\vec{q}-\vec{p})$$

Therefore
$$\vec{r}=\vec{p}+t(\vec{q}-\vec{p})$$

This represents a line passing through P and Q.
Question 9
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$$A, B, C$$ and $$D$$ have position vectors $$\vec{a}, \vec{b}, \vec{c}$$ and $$\vec{d}$$, respectively, such that $$\vec{a} - \vec{b} = 2 (\vec{d} - \vec{c})$$, then

A
$$AB$$ and $$CD$$ bisect each other

B
$$AB$$ and $$CD$$ trisect each other

C
$$BD$$ and $$AC$$ bisect each other

D
$$BD$$ and $$AC$$ trisect each other

Solution

Given that
$$\vec{a} - \vec{b} = 2 (\vec{d} - \vec{c})$$

$$\therefore \displaystyle \frac{\vec a + 2 \vec c}{2 + 1} = \frac{\vec b + 2 \vec d}{2 + 1}$$

Hence, $$AC$$ and $$BD$$ trisect each other as $$L.H.S$$ is the position vector of a point trisecting $$A$$ and $$C$$, and $$R.H.S$$. that of $$B$$ and $$D$$.
Question 10
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Let $$ABC$$ be a triangle whose centroid is $$G$$, orthocentre is $$H$$ and circumcentre is the origin '$$O$$'. If $$D$$ is any point in the plane of the triangle such that no three of $$O, A, C$$ and $$D$$ are collinear satisfying the relation $$\vec{AD} + \vec{BD} + \vec{CH} + 3 \vec{HG} = \lambda \vec{HD}$$, then what is the value of the scalar $$'\lambda'$$?

A
$$2$$

B
$$3$$

C
$$-1$$

D
$$-2$$

Solution

We know that in vector
$$\vec{AB}=\vec{OB}-\vec{OA}$$ where O is the origin
SO from ques
$$\vec{AD}+\vec{BD}+\vec{CH}+3\vec{HG}=\lambda\vec{HD}$$
$$\vec{OD}-\vec{OA}+\vec{OD}-\vec{OB}+\vec{OH}-\vec{OC}+3\vec{OG}-3\vec{OH}=\lambda\vec{HD}$$
$$2\vec{OD}-2\vec{OH}-\vec{OA}-\vec{OB}-\vec{OC}+3\vec{OG}=\lambda\vec{HD}$$
$$2(\vec{OD}-\vec{OH})-\vec{OA}-\vec{OB}-\vec{OC}+3\vec{OG}=\lambda\vec{HD}$$
$$2\vec{HD}-\vec{OA}-\vec{OB}-\vec{OC}+3\vec{OG}=\lambda\vec{HD}$$
Comparing both sides
$$\lambda=2$$

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 31

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Vector Algebra Test - 31

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SHARING IS CARING

Question 1

$$\vec{0}$$

$$3\vec{GA}$$

$$3\vec{GB}$$

$$3\vec{GC}$$

Solution

Question 2

$$\overrightarrow{AB} + \overrightarrow{BC}+\overrightarrow{CA} = \overrightarrow{0}$$

$$\overrightarrow{AB} + \overrightarrow{BC}- \overrightarrow{AC} = \overrightarrow{0}$$

$$\overrightarrow{AB} + \overrightarrow{BC}- \overrightarrow{CA} = \overrightarrow{0}$$

$$\overrightarrow{AB} - \overrightarrow{CB} + \overrightarrow{CA} = \overrightarrow{0}$$

Solution

Question 3

$$\vec b+\vec e=\vec f$$

$$\vec b+\vec c=\vec f$$

$$\vec d+\vec c=\vec f$$

$$\vec d+\vec e=\vec f$$

Solution

Question 4

$$2BC$$

$$3BC$$

$$\displaystyle \dfrac{4}{3}BC$$

None of these

Solution

Question 5

$$\vec{0}$$

$$(a+b+c) \vec{BC}$$

$$(\vec{a} + \vec{b} + \vec{c}) \vec{AC}$$

$$(a + b + c)\vec{AB}$$

Solution

Question 6

$$3$$

$$9$$

$$ \dfrac { 3 }{ 4 } $$

$$ \dfrac { 9 }{ 4 } $$

Solution

Question 7

$$\mu = \alpha + 2$$

$$\mu + \alpha =1$$

$$\alpha = \mu + 1$$

$$\mu = \alpha +1$$

Solution

Question 8

A plane containing the origin $$O$$ and parallel to two non-collinear vectors $$\vec{OP}$$ and $$\vec{OQ}$$

The surface of a sphere described on $$PQ$$ as its diameter

A line passing through points $$P$$ and $$Q$$

A set of lines parallel to line $$PQ$$

Solution

Question 9

$$AB$$ and $$CD$$ bisect each other

$$AB$$ and $$CD$$ trisect each other

$$BD$$ and $$AC$$ bisect each other

$$BD$$ and $$AC$$ trisect each other

Solution

Question 10

$$2$$

$$3$$

$$-1$$

$$-2$$

Solution

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