Vector Algebra Test

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Vector Algebra Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

If the vectors $$4\hat{i}-7\hat{j}-2\hat{k},\: \hat{i}+5\hat{j}-3\hat{k},\: 3\hat{i}-\lambda\hat{j}+\hat{k}$$ form a triangle then $$\lambda=$$

A
$$6$$

B
$$-6$$

C
$$12$$

D
$$-1$$

Solution

Triangle follow Addition Law
$$\hat{i}+5\hat{j}-3\hat{k}+3\hat{i}-\lambda \hat{j}+\hat{k}=4\hat{i}-7\hat{j}-2\hat{k}$$
$$(5-\lambda )=-7$$
$$\lambda =12$$
Question 2
1 / -0

lf $$4{\vec{i}}+7\vec{j}+8\vec{k} , 2\vec{i}+3\vec{j}+4\vec{k}$$ and $$2\vec{i}+5\vec{j}+7\vec{k}$$ are the position vectors of the vertices $$\mathrm{A},\mathrm{B}$$ and $$\mathrm{C}$$ of $$ \triangle \mathrm{A}\mathrm{B}\mathrm{C}$$, the position vector of $$D$$ the point where the bisector of $$\angle A$$ meets $$\mathrm{B}\mathrm{C}$$ is

A
$${\dfrac{2}{3}}(-6\hat {i}-8\hat {j}-6\hat {k})$$

B
$${\dfrac{2}{3}}(6\hat {i}+8\hat{j}+6\hat{k})$$

C
$${\dfrac{1}{3}}(6\hat{i}+13\hat{j}+18\hat{k})$$

D
$$2(\hat{i}+\hat{j}+\hat{k})$$

Solution
Question 3
1 / -0

lf $$\vec{a}$$ and $$\vec{b}$$ are two non-parallel unit vectors and the vector $$\alpha\vec{a}+\vec{b}$$ bisects the internal angle between $$\vec{a}$$ and $$\vec{b}$$, then $$\alpha$$ is

A
$$1$$

B
$$\dfrac{1}{2}$$

C
$$2$$

D
$$5$$

Solution

Given $$\vec{a}\ and\ \vec{b}\ are\ non-parallel\ unit\ vector$$
So
$$\left | \vec{a} \right |=1$$ and
$$\left | \vec{b} \right |=1$$
Internal angle bisector of $$\vec{a}$$ and $$\vec{b}$$ is
$$\dfrac{\vec{a}}{\left | \vec{a} \right |}+\dfrac{\vec{b}}{\left | \vec{b} \right |}$$
$$=\vec{a}+\vec{b}$$ {$$\left | \vec{a} \right |=1$$ and $$\left | \vec{b} \right |=1$$ }
and Given bisector is $$\alpha\vec{a}+\vec{b}$$
Comparing Both the bisector
$$\vec{a}+\vec{b}=\alpha\vec{a}+\vec{b}$$
from above eq
$$\alpha=1$$.
Question 4
1 / -0

lf $$A=(-3,2,5), B=(-3,4,5)$$ and $$C=(-3,4,7)$$ are the position vectors of vertices of $$\Delta ABC$$ then its circumcentre is

A
$$(-3,3,5)$$

B
$$(-3,3,6)$$

C
$$(-3,4,6)$$

D
$$(-3,4,7)$$

Solution

$$\overrightarrow{AB}=2\hat{j}$$
$$\overrightarrow{AC}=2\hat{j}+2\hat{k}$$
$$ \overrightarrow{BC}=2\hat{k}$$
Triangle $$ABC$$ is right angled at angle $$B$$.
Hence, the circumcentre is mid point of $$AC$$. Let the circumcentre be $$D$$.
$$=\frac{1}{2}(-6,6,12)$$
$$=(-3,3,6)$$
Question 5
1 / -0

$$\hat { AB } =-3{ \hat { i } }+4{ \hat { k } }$$ and $$\hat { BC } =-\overline { i } -2\overline { k } $$ are the sides of the triangle $$ ABC$$ then the length of the median $$AM$$ is

A
$$\sqrt{\dfrac{25}{2}}$$

B
$$\sqrt{\dfrac{45}{2}}$$

C
$$\displaystyle \dfrac{\sqrt{65}}{2}$$

D
$$\displaystyle \dfrac{\sqrt{85}}{2}$$

Solution

Given
$$ \overrightarrow{AB}=-3\hat{i}+4\hat{k} $$
$$ \overrightarrow{BC}=-\hat{i}-2\hat{k}$$

Let $$ \overrightarrow{A} = 0\hat{i}$$ $$+0\hat{j}$$ $$+0\hat{k} $$, $$\overrightarrow{AB} =\overrightarrow{B}-\overrightarrow{A}$$
$$\Rightarrow \overrightarrow{B} =-3\hat{i} +4\hat{k} $$
$$ \overrightarrow{BC} =\overrightarrow{C}-\overrightarrow{B} $$
$$ \overrightarrow{C} = -3\hat{i}+4\hat{k}+(-\hat{i}-2\hat{k})$$, $$\overrightarrow{C} = -4\hat{i}+2\hat{k} $$
Median $$\overrightarrow{AM} = \dfrac{\overrightarrow{B}+\overrightarrow{C}}{2}$$
$$\overrightarrow{AM}= -\dfrac{7}{2}\hat{i} + 3\hat{k}$$

$$\left | \overrightarrow{AM} \right | =\sqrt{(\dfrac{7}{2})^2+3^2}$$

$$\left | \overrightarrow{AM} \right | =\sqrt{\dfrac{85}{4}}$$

$$ \left | \overrightarrow{AM} \right | =\dfrac{\sqrt{85}}{2}$$
Hence, option D is correct.
Question 6
1 / -0

Two forces act at the vertex $$\mathrm{A}$$ of quadrilateral $$\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$$ represented by $$\overline{AB},\ \overline{AD}$$ and two at $$\mathrm{C}$$ represented by $$\overline{CD}$$ and $$\overline{CB}$$. If $$\mathrm{E},\ \mathrm{F}$$ are mid points of $$\overline{AC}$$ and $$\overline{BD}$$ respectively, then their resultant is

A
$$\overline{EF}$$

B
$$2\overline{EF}$$

C
$${\dfrac{3}{2}\overline{EF}}$$

D
$$4\overline{EF}$$

Solution

Let position vector of $$A,B,C,D$$ are $$\vec{a},\vec{b},\vec{c},\vec{d}$$
Position Vector of $$E=\dfrac{\vec{a}+\vec{c}}{2}$$
Position Vector of $$F=\dfrac{\vec{b}+\vec{d}}{2}$$
So, $$\vec{AB}=\vec{b}-\vec{a}$$, $$\vec{AD}=\vec{d}-\vec{a}$$, $$\vec{CD}=\vec{d}-\vec{c}$$ and $$\vec{CB}=\vec{b}-\vec{c}$$
Resultant=$$\vec{AB}+\vec{AD}+\vec{CD}+\vec{CB}$$
$$=2(\vec{b}-a+\vec{d}-\vec{c})$$
$$=2(2F-2E)$$
$$=4\vec{FE}$$
Question 7
1 / -0

If point $$O$$ is the centre of a circle circumscribed about a triangle $$ABC$$. Then $$\overline{OA}\sin 2A+\overline{OB}\sin2B+\overline{OC}\sin 2C=$$

A
$$(\overline{OA}+\overline{OB}+\overline{OC})\sin 2A$$

B
$$(\overline{OA}+\overline{OB}+\overline{OC})\cos 2A$$

C
$$\overline{0}$$

D
$$(\overline{OA}+\overline{OB}+\overline{OC})\tan 2A$$

Solution

In order to make the problem easier and take less time to solve it, we take a simpler case and take the triangle to be an equilateral one.
Thus $$OA = OB = OC$$ and $$sin 2A = sin 2B = sin 2C = sin (120^0) = \dfrac{\sqrt {3}}{2}$$
So after solving we get $$\overline {OA}$$ + $$\overline{OB} $$ + $$\overline{OC}$$ $$= 0.$$
Hence the answer comes out to be $$0.$$
Question 8
1 / -0

Let $$G$$ and $$G^{1}$$ be the centroids of the triangles $$ABC$$ and $$A^{1}B^{1}C^{1}$$ respectively, then $$AA^{1}+BB^{1}+CC^{1}$$ is equal to

A
$$2GG^{1}$$

B
$$3G^{1}G$$

C
$$3GG^{1}$$

D
$$\displaystyle \dfrac{3}{2}GG^{1}$$

Solution

$$G=\dfrac{A+B+C}{3}\Rightarrow A+B+C=3G$$ ......(1)
and $$G^1=\dfrac{A^1+B^1+C^1}{3}\Rightarrow A^1+B^1+C^1=3G^1$$ ......(2)
$$\therefore AA^1+BB^1+CC^1 = (A-A^1)+(B-B^1)+(C-C^1)$$
$$=(A+B+C)-(A^1+B^1+C^1)$$
$$= 3G-3G^1=3(G-G^1)=3GG^1$$, Using (1) and (2)
Question 9
1 / -0

The ratio in which $$\overline{i}+2\overline{j}+3\overline{k}$$ divides the join $$\mathrm{o}\mathrm{f}-2\overline{i}+3\overline{j}+5\overline{k}$$ and $$7\overline{i}-\overline{k}$$ is

A
$$-3:2$$

B
$$1:2$$

C
$$2:3$$

D
$$-4:3$$

Solution

Given
$$ i+j+k$$ divides $$ -2\overline{ i } +3\overline { j } +5\overline { k }, 7\overline { i } -\overline { k }$$
$$ { x }_{ 1 }=-2, { x }_{ 2 }=1, { x }_{ 3 }=7$$
Now, Required Ratio$$ = { x }_{ 1 }-{ x }_{ 2 }: { x }_{ 2 }-{ x }_{ 3 }$$(or) $${ y }_{ 1 }-{ y }_{ 2 } : { y }_{ 2 }-{ y }_{ 3 } =-2-(1) : 1-7 = -3: -6 = 1:2$$
OPTION B
Question 10
1 / -0

Orthocentre of an equilateral triangle $$ABC$$ is the origin $$\mathrm{O}$$. If $$A=\overline{a},\ B=\overline{b},\ C=\overline{c}$$ then $$\overline{AB}+2\overline{BC}+3\overline{CA}=$$

A
$$3\overline{c}$$

B
$$3\overline{a}$$

C
$$0$$

D
$$3\overline{b}$$

Solution

$$\overline{AB}+2\overline{BC}+3\overline{CA}$$
$$=\vec{b}-\vec{a}+2(\vec{c}-\vec{b})+3(\vec{a}-\vec{c})$$
$$=-\vec{b}+2\vec{a}- \vec{c}$$ ----------(1)

orthocentre = circumcenter (property of equalatoral $$\Delta $$)
so circumecentre $$= \dfrac{\vec{a}+\vec{b}+\vec{c}}{3}$$
$$0= \dfrac{\vec{a}+\vec{b}+\vec{c}}{3}$$
$$-\vec{b}=\vec{a}+\vec{c}$$
so put in (1)
$$3\vec{a}$$

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 25

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Vector Algebra Test - 25

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SHARING IS CARING

Question 1

$$6$$

$$-6$$

$$12$$

$$-1$$

Solution

Question 2

$${\dfrac{2}{3}}(-6\hat {i}-8\hat {j}-6\hat {k})$$

$${\dfrac{2}{3}}(6\hat {i}+8\hat{j}+6\hat{k})$$

$${\dfrac{1}{3}}(6\hat{i}+13\hat{j}+18\hat{k})$$

$$2(\hat{i}+\hat{j}+\hat{k})$$

Solution

Question 3

$$1$$

$$\dfrac{1}{2}$$

$$2$$

$$5$$

Solution

Question 4

$$(-3,3,5)$$

$$(-3,3,6)$$

$$(-3,4,6)$$

$$(-3,4,7)$$

Solution

Question 5

$$\sqrt{\dfrac{25}{2}}$$

$$\sqrt{\dfrac{45}{2}}$$

$$\displaystyle \dfrac{\sqrt{65}}{2}$$

$$\displaystyle \dfrac{\sqrt{85}}{2}$$

Solution

Question 6

$$\overline{EF}$$

$$2\overline{EF}$$

$${\dfrac{3}{2}\overline{EF}}$$

$$4\overline{EF}$$

Solution

Question 7

$$(\overline{OA}+\overline{OB}+\overline{OC})\sin 2A$$

$$(\overline{OA}+\overline{OB}+\overline{OC})\cos 2A$$

$$\overline{0}$$

$$(\overline{OA}+\overline{OB}+\overline{OC})\tan 2A$$

Solution

Question 8

$$2GG^{1}$$

$$3G^{1}G$$

$$3GG^{1}$$

$$\displaystyle \dfrac{3}{2}GG^{1}$$

Solution

Question 9

$$-3:2$$

$$1:2$$

$$2:3$$

$$-4:3$$

Solution

Question 10

$$3\overline{c}$$

$$3\overline{a}$$

$$0$$

$$3\overline{b}$$

Solution

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