Vector Algebra Test

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Vector Algebra Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

If $\bar{\alpha }$ , $\bar{\beta }$ and $\bar{\gamma }$ be vertices of a $\triangle$ whose circumcenter is at the origin, then orthocenter is given by

A
$\displaystyle \dfrac{\bar{\alpha }+\bar{\beta }+\bar{\gamma }}{4}$

B
$\displaystyle \dfrac{\bar{\alpha }+\bar{\beta }+\bar{\gamma }}{2}$

C
$\bar{\alpha }+\bar{\beta }+\bar{\gamma }$

D
$\displaystyle \dfrac{\bar{\alpha }+\bar{\beta }+\bar{\gamma }}{3}$

Solution

We have $\bar\alpha$ , $\bar\beta$ , $\bar\gamma$ vertices of a triangle.
Centroid of the $\triangle$ is $\displaystyle \dfrac{\bar{\alpha }+\bar{\beta }+\bar{\gamma }}{3}$
Now centroid divides the line joining orthocenter and circumcenter in ratio $2:1$
$\therefore$ orthocenter of the $\triangle$ is $\bar{\alpha }+\bar{\beta }+\bar{\gamma }$
Question 2
1 / -0

If $S$ is the circumcenter, $O$ is the orthocenter of $\triangle ABC$ , then $\displaystyle \vec{SA}+\vec{SB}+\vec{SC}=$

A
$2\vec{OS}$

B
$2\vec{SO}$

C
$\vec{OS}$

D
$\vec{SO}$

Solution

A very important property to be remembered of a triangle is that the centroid divides the line joining orthocentre and circumcentre in the ratio $2 : 1$
$\Rightarrow C = \dfrac{2S + O}{3}$

$\Rightarrow\dfrac{\overline{A} + \overline{B} + \overline{C}}{3} = \dfrac{2\overline{S} + \overline{O}}{3}$

$\Rightarrow \overline{A} + \overline{B} + \overline{C} = 2\overline{S} + \overline{O}$

Thus, $\overline{SA} + \overline{SB} + \overline{SC} = \overline{A} + \overline{B} + \overline{C} - 3\overline{S} = \overline{SO}$
Question 3
1 / -0

For non-zero vectors $\bar{a}$ , $\bar{b}$ and $\bar{c}$ , $\left | \left ( \bar{a}\times \bar{b} \right ).\bar{c} \right |=\left | \bar{a} \right |\left | \bar{b} \right |\left | \bar{c} \right |$ iff

A
$\bar{a}.\bar{c}=0$ , $\bar{a}.\bar{b}=0$

B
$\bar{a}.\bar{b}=0$ , $\bar{b}.\bar{c}=0$

C
$\bar{c}.\bar{a}=0$ , $\bar{b}.\bar{c}=0$

D
$\bar{a}.\bar{b}=\bar{b}.\bar{c}=\bar{c}.\bar{a}=0$

Solution

$\left | \left ( \overrightarrow{a}\times \overrightarrow{b} \right ).\overrightarrow{c} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\left | \overrightarrow{c} \right |$ $(\theta$ is angle between $\overrightarrow{c}$ and plane of $\overrightarrow{a}\times \overrightarrow{b})$

$\Rightarrow \overrightarrow{c}$ is $\perp$ to $\overrightarrow{a}$ and $\overrightarrow{b}$ for $\cos \theta =1$

and $\left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |$ iff $\overrightarrow{a}\perp \overrightarrow{b}$

$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}=0=\overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{c}\cdot \overrightarrow{a}$
Question 4
1 / -0

Two planes are perpendicular to each other,one of them contains vector $\bar{a}$ and $\bar{b}$ , other contains $\bar{c}$ and $\bar{d}$ then $\left ( \bar{a}\times \bar{b} \right )\cdot \left (\bar{c}\times \bar{d} \right )=$

A
$1$

B
$0$

C
$\left [ \bar{a}\bar{b}\bar{c} \right ]$

D
$\left [ \bar{b}\bar{c}\bar{d} \right ]$

Solution

Normal vector to plane $\bar{a}, \bar{b}$ is given by, $\bar{a}\times \bar{b}$
And normal vector to plane $\bar{c}, \bar{d}$ is given by, $\bar{c}\times \bar{d}$
Now for planes to be pependicular, the angle between their normal should be $90^{\circ}$
$\Rightarrow (\bar{a}\times \bar{b})\cdot (\bar{c}\times \bar{d}) = |\bar{a}\times \bar{b}||\bar{c}\times \bar{d}|\cos 90^{\circ} = 0$
Question 5
1 / -0

In a parallelogram $ABCD,\left| AB \right| =a,\left| AD \right| =b$ and $\left| AC \right| =c.$ Then, $DB.AB$ has the value

A
$\displaystyle\frac { 3{ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2 }$

B
$\displaystyle\frac { { a }^{ 2 }+3{ b }^{ 2 }-{ c }^{ 2 } }{ 2 }$

C
$\displaystyle\frac { { a }^{ 2 }-{ b }^{ 2 }+3{ c }^{ 2 } }{ 2 }$

D
$\displaystyle\frac { { a }^{ 2 }+3{ b }^{ 2 }+{ c }^{ 2 } }{ 2 }$

Solution

$\because DB=DA+AB$ or, $DA=DB-AB$
$\displaystyle \therefore { \left( DA \right) }^{ 2 }={ \left( DB \right) }^{ 2 }+{ \left( AB \right) }^{ 2 }-2DB.AB$
In parallelogram $\displaystyle 2\left( { a }^{ 2 }+{ b }^{ 2 } \right) ={ c }^{ 2 }+{ DB }^{ 2 }$
$\displaystyle \therefore { \left( DB \right) }^{ 2 }=2{ a }^{ 2 }+2{ b }^{ 2 }-{ c }^{ 2 }$
$\therefore$ From (1), $\displaystyle { b }^{ 2 }=2{ a }^{ 2 }+{ 2b }^{ 2 }-{ c }^{ 2 }+{ a }^{ 2 }-2AB.DB$
$\displaystyle \therefore AB.DB=\dfrac { 3{ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2 }$ .
Question 6
1 / -0

If a. $\displaystyle b\neq 0,$ find the vector r which satisfies the equations $\displaystyle \left ( r-c \right )\times b= 0, r.a= 0$

A
$\displaystyle r= \frac{\left ( a.b \right )c+\left ( a.c \right )b}{a.b}$

B
$\displaystyle r= \frac{\left ( a.b \right )c-\left ( a.c \right )b}{a.b}$

C
$\displaystyle r= \frac{-\left ( a.b \right )c-\left ( a.c \right )b}{a.b}$

D
$\displaystyle r= \frac{-\left ( a.b \right )c+\left ( a.c \right )b}{a.b}$

Solution

$\displaystyle \left ( r-c \right )\times b= 0\Rightarrow r-c$ and b are collinear
$\displaystyle \therefore r-c= kb$ or $\displaystyle r= c+kb$ ...(1)
Again $\displaystyle r.a=0 \therefore \left ( c+kb \right ).a= 0$ or
$\displaystyle c.a+kb.a= 0 \therefore k= -\dfrac{a.c}{a.b}$
Substitute in (1), we get $\displaystyle r= c-\left ( \dfrac{a.c}{a.b} \right )b$
Hence $\displaystyle r= \dfrac{\left ( a.b \right )c-\left ( a.c \right )b}{a.b}$
Question 7
1 / -0

If $\alpha \bar{a}+\beta \bar{b}+\gamma \bar{c}=0$ , then $\left ( \bar{a}\times \bar{b} \right )\times \left [ \left ( \bar{b}\times \bar{c} \right )\times \left ( \bar{c}\times \bar{a} \right ) \right ]$ is equal to

A
$\bar{0}$

B
A vector $\perp$ plane of $\bar{a}$ , $\bar{b}$ and $\bar{c}$

C
A scalar quantity

D
None of these

Solution

Since $\alpha \bar{a}+\beta \bar{b}+\gamma \bar{c}=0$
$\therefore$ the vectors $\bar{a}$ , $\bar{b}$ and $\bar{c}$ are coplanar
and thus $\left ( \bar{c}\times \bar{b} \right )$ and $\left ( \bar{c}\times \bar{a} \right )$ are coplanar.
$\therefore$ $\left ( \bar{b}\times \bar{c} \right )\times \left ( \bar{c}\times \bar{a} \right )=0$
Hence option $A$ becomes correct.
Question 8
1 / -0

The position vector of the points $A$ and $B$ are respectively $\bar{a}$ and $\bar{b}$ divides $AB$ in the ratio $3:1$ and $Q$ iis the midpoint of $AP$ . The position vector of $Q$ is

A
$\displaystyle \dfrac{3\bar{a}+5\bar{b}}{8}$

B
$\displaystyle\dfrac{3\bar{a}+\bar{b}}{4}$

C
$\displaystyle\dfrac{\bar{a}+3\bar{b}}{4}$

D
$\displaystyle\dfrac{5\bar{a}+3\bar{b}}{8}$

Solution

Given
position vector of A
$\vec{OA}=\vec{a}$
position vector of B
$\vec{OB}=\vec{b}$
Point P divides AB in ration $3:1$
position vector of Point P
$\vec{OP}=\dfrac { 3\bar { OB } +\bar { OA } }{ 4 }$
$\vec{OP}=\dfrac { 3\bar { b } +\bar { a} }{ 4 }$
Here Q is the mid point of AP
SO $\vec{OQ}=\dfrac{\vec{OA}+\vec{OP}}{2}$
$Q=\dfrac { \dfrac { 3\bar { b } +\bar { a } }{ 4 } +\bar { a } }{ 2 } =\dfrac { 5\bar { a } +3\bar { b } }{ 8 }$
Question 9
1 / -0

For three unit vectors $\bar{a}$ , $\bar{b}$ and $\bar{c}$ , if $\bar{a}+\bar{b}+\bar{c}= \bar{0}$ , then the value of $\bar{a}\cdot \left ( \bar{b}+\bar{c} \right )+\bar{b}\cdot \left ( \bar{c}+\bar{a} \right )+\bar{c}\cdot \left ( \bar{a}+\bar{b} \right )$ is equal to

A
$-\dfrac{3}{2}$

B
$0$

C
$-3$

D
None of these

Solution

Given $\bar{a} + \bar{b} + \bar{c} = 0$
We have
$\therefore\bar{a}\cdot \left ( \bar{b}+\bar{c} \right )+\bar{b}\cdot \left ( \bar{c}+\bar{a} \right )+\bar{c}\cdot \left ( \bar{a}+\bar{b} \right )$
$= \bar{a}. - \bar{a} - \bar{b}.\bar{b} - \bar{c}.\bar{c}$
$= -a^2 - b^2 - c^2$
If all the three vectors are unit vectors, the answer will be $-3$
Question 10
1 / -0

Let ABCD be a parallelogram whose diagonals intersect at P and let O be the origin, then $\bar{OA}+\bar{OB}+\bar{OC}+\bar{OD}=$

A
$2\bar{OP}$

B
$3\bar{OP}$

C
$\bar{OP}$

D
$4\bar{OP}$

Solution

Let the position vectors of $A, B, C, D, P$ be $\overline{a}, \overline{b}, \overline{c}, \overline{d}, \overline{p}$ respectively.
We need $\overline{a} + \overline{b} + \overline{c} + \overline{d}$
Also, $\overline{p} = \dfrac{\overline{a} + \overline{b}}{2} = \dfrac{\overline{c} + \overline{d}}{2}$ ......... $(\because$ diagonals bisect each other $)$
So, the question needed becomes $4\overline{p} = 4\overline{OP}$

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Vector Algebra Test - 35

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Vector Algebra Test - 35

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Question 1

αˉ+βˉ+γˉ4\displaystyle \dfrac{\bar{\alpha }+\bar{\beta }+\bar{\gamma }}{4}4αˉ+βˉ​+γˉ​​

αˉ+βˉ+γˉ2\displaystyle \dfrac{\bar{\alpha }+\bar{\beta }+\bar{\gamma }}{2}2αˉ+βˉ​+γˉ​​

αˉ+βˉ+γˉ\bar{\alpha }+\bar{\beta }+\bar{\gamma }αˉ+βˉ​+γˉ​

αˉ+βˉ+γˉ3\displaystyle \dfrac{\bar{\alpha }+\bar{\beta }+\bar{\gamma }}{3}3αˉ+βˉ​+γˉ​​

Solution

Question 2

2OS⃗2\vec{OS}2OS

2SO⃗2\vec{SO}2SO

OS⃗\vec{OS}OS

SO⃗\vec{SO}SO

Solution

Question 3

aˉ.cˉ=0\bar{a}.\bar{c}=0aˉ.cˉ=0, aˉ.bˉ=0\bar{a}.\bar{b}=0aˉ.bˉ=0

aˉ.bˉ=0\bar{a}.\bar{b}=0aˉ.bˉ=0, bˉ.cˉ=0\bar{b}.\bar{c}=0bˉ.cˉ=0

cˉ.aˉ=0\bar{c}.\bar{a}=0cˉ.aˉ=0, bˉ.cˉ=0\bar{b}.\bar{c}=0bˉ.cˉ=0

aˉ.bˉ=bˉ.cˉ=cˉ.aˉ=0\bar{a}.\bar{b}=\bar{b}.\bar{c}=\bar{c}.\bar{a}=0aˉ.bˉ=bˉ.cˉ=cˉ.aˉ=0

Solution

Question 4

111

000

[aˉbˉcˉ]\left [ \bar{a}\bar{b}\bar{c} \right ][aˉbˉcˉ]

[bˉcˉdˉ]\left [ \bar{b}\bar{c}\bar{d} \right ] [bˉcˉdˉ]

Solution

Question 5

3a2+b2−c22\displaystyle\frac { 3{ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2 } 23a2+b2−c2​

a2+3b2−c22\displaystyle\frac { { a }^{ 2 }+3{ b }^{ 2 }-{ c }^{ 2 } }{ 2 } 2a2+3b2−c2​

a2−b2+3c22\displaystyle\frac { { a }^{ 2 }-{ b }^{ 2 }+3{ c }^{ 2 } }{ 2 } 2a2−b2+3c2​

a2+3b2+c22\displaystyle\frac { { a }^{ 2 }+3{ b }^{ 2 }+{ c }^{ 2 } }{ 2 } 2a2+3b2+c2​

Solution

Question 6

r=(a.b)c+(a.c)ba.b\displaystyle r= \frac{\left ( a.b \right )c+\left ( a.c \right )b}{a.b}r=a.b(a.b)c+(a.c)b​

r=(a.b)c−(a.c)ba.b\displaystyle r= \frac{\left ( a.b \right )c-\left ( a.c \right )b}{a.b}r=a.b(a.b)c−(a.c)b​

r=−(a.b)c−(a.c)ba.b\displaystyle r= \frac{-\left ( a.b \right )c-\left ( a.c \right )b}{a.b}r=a.b−(a.b)c−(a.c)b​

r=−(a.b)c+(a.c)ba.b\displaystyle r= \frac{-\left ( a.b \right )c+\left ( a.c \right )b}{a.b}r=a.b−(a.b)c+(a.c)b​

Solution

Question 7

0ˉ\bar{0}0ˉ

A vector ⊥\perp ⊥ plane of aˉ\bar{a}aˉ, bˉ\bar{b}bˉ and cˉ\bar{c}cˉ

A scalar quantity

None of these

Solution

Question 8

3aˉ+5bˉ8\displaystyle \dfrac{3\bar{a}+5\bar{b}}{8}83aˉ+5bˉ​

3aˉ+bˉ4\displaystyle\dfrac{3\bar{a}+\bar{b}}{4}43aˉ+bˉ​

aˉ+3bˉ4\displaystyle\dfrac{\bar{a}+3\bar{b}}{4}4aˉ+3bˉ​

5aˉ+3bˉ8\displaystyle\dfrac{5\bar{a}+3\bar{b}}{8}85aˉ+3bˉ​

Solution

Question 9

−32-\dfrac{3}{2}−23​

000

−3-3−3

None of these

Solution

Question 10

2OPˉ2\bar{OP} 2OPˉ

3OPˉ3\bar{OP} 3OPˉ

OPˉ\bar{OP} OPˉ

4OPˉ4\bar{OP}4OPˉ

Solution

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$\displaystyle \dfrac{\bar{\alpha }+\bar{\beta }+\bar{\gamma }}{4}$

$\displaystyle \dfrac{\bar{\alpha }+\bar{\beta }+\bar{\gamma }}{2}$

$\bar{\alpha }+\bar{\beta }+\bar{\gamma }$

$\displaystyle \dfrac{\bar{\alpha }+\bar{\beta }+\bar{\gamma }}{3}$

$2\vec{OS}$

$2\vec{SO}$

$\vec{OS}$

$\vec{SO}$

$\bar{a}.\bar{c}=0$ , $\bar{a}.\bar{b}=0$

$\bar{a}.\bar{b}=0$ , $\bar{b}.\bar{c}=0$

$\bar{c}.\bar{a}=0$ , $\bar{b}.\bar{c}=0$

$\bar{a}.\bar{b}=\bar{b}.\bar{c}=\bar{c}.\bar{a}=0$

$1$

$0$

$\left [ \bar{a}\bar{b}\bar{c} \right ]$

$\left [ \bar{b}\bar{c}\bar{d} \right ]$

$\displaystyle\frac { 3{ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2 }$

$\displaystyle\frac { { a }^{ 2 }+3{ b }^{ 2 }-{ c }^{ 2 } }{ 2 }$

$\displaystyle\frac { { a }^{ 2 }-{ b }^{ 2 }+3{ c }^{ 2 } }{ 2 }$

$\displaystyle\frac { { a }^{ 2 }+3{ b }^{ 2 }+{ c }^{ 2 } }{ 2 }$

$\displaystyle r= \frac{\left ( a.b \right )c+\left ( a.c \right )b}{a.b}$

$\displaystyle r= \frac{\left ( a.b \right )c-\left ( a.c \right )b}{a.b}$

$\displaystyle r= \frac{-\left ( a.b \right )c-\left ( a.c \right )b}{a.b}$

$\displaystyle r= \frac{-\left ( a.b \right )c+\left ( a.c \right )b}{a.b}$

$\bar{0}$

A vector $\perp$ plane of $\bar{a}$ , $\bar{b}$ and $\bar{c}$

$\displaystyle \dfrac{3\bar{a}+5\bar{b}}{8}$

$\displaystyle\dfrac{3\bar{a}+\bar{b}}{4}$

$\displaystyle\dfrac{\bar{a}+3\bar{b}}{4}$

$\displaystyle\dfrac{5\bar{a}+3\bar{b}}{8}$

$-\dfrac{3}{2}$

$0$

$-3$

$2\bar{OP}$

$3\bar{OP}$

$\bar{OP}$

$4\bar{OP}$