Vector Algebra Test

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Vector Algebra Test - 35

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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
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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
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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
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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
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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
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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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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 35

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Vector Algebra Test - 35

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SHARING IS CARING

Question 1

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

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

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

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

Solution

Question 2

$$2\vec{OS}$$

$$2\vec{SO}$$

$$\vec{OS}$$

$$\vec{SO}$$

Solution

Question 3

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

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

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

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

Solution

Question 4

$$1$$

$$0$$

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

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

Solution

Question 5

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

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

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

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

Solution

Question 6

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

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

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

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

Solution

Question 7

$$\bar{0}$$

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

A scalar quantity

None of these

Solution

Question 8

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

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

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

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

Solution

Question 9

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

$$0$$

$$-3$$

None of these

Solution

Question 10

$$2\bar{OP} $$

$$3\bar{OP} $$

$$\bar{OP} $$

$$4\bar{OP}$$

Solution

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