Vector Algebra Test

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Vector Algebra Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

What is the maximum number of components into which a vector can be
split ?

A
2

B
3

C
4

D
Infinite

Solution

A vector can be split into infinite components (but only 3 orthogonal ones)
Question 2
1 / -0

$$P$$ is any point on the circumcircle of $$\triangle ABC$$ other than the vertices. $$H$$ is the orthocenter of $$\triangle ABC,M$$ is the mid-point of $$PH$$ and $$D$$ is the mid-point of $$BC$$. Then

A
$$AP$$ is opposite side of $$DM$$

B
$$DM$$ is parallel to $$AP$$

C
$$DM$$ is perpendicular to $$AP$$

D
None of these

Solution

Let $$S$$ be the circumcentre of $$\triangle ABC.$$
Let $$\displaystyle \overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } $$ and $$\overrightarrow { p } $$ be the orthocentre of vectors of $$A,B,C$$ and $$P$$ respectively with referentre to origin $$S.$$
$$\displaystyle \overrightarrow { SA } =\overrightarrow { a } ,\overrightarrow { SB } =\overrightarrow { b } ,\overrightarrow { SC } =\overrightarrow { c } ,\overrightarrow { SP } =\overrightarrow { p } $$ and $$\displaystyle \left| \overrightarrow { p } \right| =\left| \overrightarrow { a } \right| =\left| \overrightarrow { b } \right| =\left| \overrightarrow { c } \right| =R=\\ $$ circumradius.
Now, $$\displaystyle \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =\overrightarrow { SA } +\overrightarrow { SB } +\overrightarrow { SC } $$
$$\displaystyle =\overrightarrow { SA } +\overrightarrow { 2SD } =\overrightarrow { SA } +\overrightarrow { AH } =\overrightarrow { SH } $$
$$(\because H.G:GS=2:1$$ where $$G$$ is centroid $$)$$
Position vector of $$\displaystyle H=\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } $$
So, position vector of $$\displaystyle M=\frac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } +\overrightarrow { p } }{ 2 } $$
$$\overrightarrow { DM } =p.v$$ of $$M-p.v.$$ of $$\displaystyle D=\frac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } +\overrightarrow { p } }{ 2 } -\frac { \overrightarrow { b } +\overrightarrow { c } }{ 2 } =\frac { \overrightarrow { a } +\overrightarrow { p } }{ 2 } $$
Now $$\displaystyle \overrightarrow { DM } .\overrightarrow { PA } =\left( \frac { \overrightarrow { a } +\overrightarrow { p } }{ 2 } \right) .\left( \frac { \overrightarrow { a } -\overrightarrow { p } }{ 2 } \right) =\frac { { \left| \overrightarrow { a } \right| }^{ 2 }-{ \left| \overrightarrow { p } \right| }^{ 2 } }{ 2 } =0\Rightarrow \overrightarrow { DM } \bot \overrightarrow { AP } $$
Question 3
1 / -0

If $$\vec{a}=(\lambda\,x)\hat{i}+(y)\hat{j}+(4z)\hat{k},\,\vec{b}=y\hat{i}+x\hat{j}+3y\hat{k},\,\vec{c}=-z\hat{i}-2z\hat{j}-\begin{pmatrix}(\lambda+1)x\end{pmatrix}\hat{k}$$ are sides of triangle as shown in figure, then value of $$\lambda$$ is (where $$x,\,y,\,z$$ are not all zero)

A
$$0$$

B
$$2$$

C
$$-1$$

D
$$1$$

Solution

$$\vec{a}+\vec{b}=\vec{c}$$
$$((\lambda\,x)\hat{i}+y\hat{j}+4z\hat{k})+(y\hat{i}+x\hat{j}+3y\hat{k})$$
$$=-z\hat{i}-2z\hat{j}-(\lambda+1)x\hat{k}$$
$$\Rightarrow\;\lambda\,x+y+z=0;\;\;x+y+2z=0$$
$$\&\;(\lambda+1)x+3y+4z=0$$
For non-trivial solution,
$$\begin{vmatrix}\lambda&1&1\\1&1&2\\(\lambda+1) & 3 & 4\end{vmatrix}=0$$
$$\Rightarrow \lambda(4-6)-(4-2(\lambda+1))+(3-(\lambda+1))=0$$
$$\Rightarrow -2\lambda-4+2\lambda+2+3-\lambda-1=0$$
$$\Rightarrow -\lambda=0\;\;\;$$
$$\Rightarrow\;\;\;\lambda=0$$
Question 4
1 / -0

For any vector a, the value of $$\displaystyle { \left( a\times i \right) }^{ 2}+{ \left( a\times k \right) }^{ 2 }$$ is equal to

A
$$\displaystyle 3{ a }^{ \wedge }2$$

B
$$\displaystyle { a }^{ \wedge }2$$

C
$$\left( {a_1^2 + 2a_2^2 + a_3^2} \right)$$

D
$$\displaystyle 4{ a }^{ \wedge }2$$

E
Answer Required

Solution

$$\begin{array}{l} { \left( { a\times i } \right) ^{ 2 } }+{ \left( { a\times k } \right) ^{ 2 } } \\ Let \\ \vec { a } ={ a_{ 1 } }\hat { i } +{ a_{ 2 } }\hat { j } +{ a_{ 3 } }\hat { k } \\ ={ \left\{ { \left( { { a_{ 1 } }\hat { i } +{ a_{ 2 } }\hat { j } +{ a_{ 3 } }\hat { k } } \right) \times \hat { i } } \right\} ^{ 2 } }+{ \left\{ { \left( { { a_{ 1 } }\hat { i } +{ a_{ 2 } }\hat { j } +{ a_{ 3 } }\hat { k } } \right) \times \hat { k } } \right\} ^{ 2 } } \\ ={ \left( { { a_{ 3 } }\hat { i } -{ a_{ 2 } }\hat { k } } \right) ^{ 2 } }+{ \left( { { a_{ 2 } }\hat { i } -{ a_{ 1 } }\hat { j } } \right) ^{ 2 } } \\ =a_{ 3 }^{ 2 }+a_{ 2 }^{ 2 }+a_{ 2 }^{ 2 }+a_{ 1 }^{ 2 } \\ =\left( { a_{ 1 }^{ 2 }+2a_{ 2 }^{ 2 }+a_{ 3 }^{ 2 } } \right) \\ Hence, \\ option\, \, C\, \, is\, correct\, \, answer. \end{array}$$
Question 5
1 / -0

If $$S$$ is the circumcentre, $$O$$ is the orthocenter of $$\triangle ABC, $$ then $$SA+SB+SC=$$

A
$$SO$$

B
$$2SO$$

C
$$OS$$

D
$$2OS$$

Solution

From geometry
$$2SD=AO,$$ where $$D$$ is the mid.
Point of $$BC.$$
$$\displaystyle \therefore SA+SB+SC=SA+SD+DB+SD+DC$$
$$=SA+2SD$$
$$=SA+AO=SO$$
$$\displaystyle \therefore SA+SB+SC=SO.$$
Question 6
1 / -0

Any vector in an arbitrary direction can always be replaced by two (or three)

A
parallel vectors which have the original vector as their resultant.

B
mutually perpendicular vectors which have the original vector as their resultant.

C
arbitrary vectors which have the original vector as their resultant.

D
it is not possible to resolve a vector.

Solution

Any vector in an arbitrary direction can always be replaced by two or three arbitrary vector because they together produce the same effect as is produced by the given vector.
Question 7
1 / -0

$$12$$ coplanar non collinear forces (all of equal magnitude) maintain a body in equilibrium, then the angle between any two adjacent forces is:

A
$$\;15^{\circ}$$

B
$$\;30^{\circ}$$

C
$$\;45^{\circ}$$

D
$$\;60^{\circ}$$

Solution

According to polygon law of vector addition, the 11 forces will be the sides of polygon in same order in direction and 1 forces will represent the resultant which is a side in opposite order of the polygon. Total angle enclosed by polygon is $$360^o$$. So, the required angle $$=\dfrac{360^o}{12}=30^o$$
Question 8
1 / -0

There are three points with position vectors $$-2a+3b+5c, a+2b+3c$$ and $$7a-c$$. What is the relation between the three points

A
Collinear

B
Forms a triangle

C
In different plane

D
None of these

Solution

Let
$$A,B,C$$ are three vectors whose position vectors are $$7\vec a - \vec c,\vec a + 2\vec b + 3\vec c, - 2\vec a + 3\vec b + 5\vec c$$ respectively.
Then
the vectors $$\vec {AB},\vec {BC}, \vec {CA}$$ are given by

$$\vec {AB}=\vec {OB}-\vec {OA}$$

$$=(\vec a++2\vec b+3\vec c)-(7\vec a-\vec c)$$

$$=-6\vec a+2\vec b+4\vec c$$

$$\vec {BC}=\vec (OC)-\vec {OB}$$

$$=(-2\vec a+3\vec b+5\vec c)-(\vec a+2\vec b+3\vec c)$$

$$=-3\vec a+\vec b+2\vec c$$

$$\vec {CA}=\vec {OC}-\vec {OA}$$

$$=(-2\vec a+3\vec b+5\vec c)-(7\vec a-\vec c)$$

$$=-9\vec a+3\vec b+6\vec c$$
And
$$\vec {AB}+\vec {BC}=(-6\vec a+2\vec b+4\vec c)+(-3\vec a+\vec b+2\vec c)$$

$$=-9\vec a+3\vec b+6\vec c$$

$$=\vec {CA}$$This implies that the points $$A,B,C$$ are collinear .
Question 9
1 / -0

If $$\vec { a } \cdot \hat { i } =4$$, then $$\left( \vec { a } \times \hat { j } \right) \cdot \left( 2\hat { j } -3\hat { k } \right) $$ is equal to

A
$$12$$

B
$$2$$

C
$$0$$

D
$$-12$$

Solution

$$\left( \vec { a } \times \hat { j } \right) \cdot \left( 2\hat { j } -3\hat { k } \right) =\vec { a } \cdot \left\{ \hat { j } \times \left( 2\hat { j } -3\hat { k } \right) \right\} $$
$$=\vec { a } \cdot \left\{ -3\left( \hat { j } \times \hat { k } \right) \right\} $$
$$=-3\left( \vec { a } \cdot \hat { i } \right) $$
$$=-12$$ ($$\because \vec { a } \cdot \hat { i } =4$$ given)
Question 10
1 / -0

If the position vectors of the vertices of a triangle be 6i + 4j + 5k, 4i + 5j + 6k and 5i + 6j + 4k then the triangle is

A
Right angled triangle

B
Equilateral triangle

C
Isosceles triangle

D
None of these

Solution

We have,

Let the vertices of triangle is

$$ \overrightarrow{OA}=\left( 6\overrightarrow{i}+4\overrightarrow{j}+5\overrightarrow{k} \right) $$

$$ \overrightarrow{OB}=\left( 4\overrightarrow{i}+5\overrightarrow{j}+6\overrightarrow{k} \right) $$

$$ \overrightarrow{OC}=\left( 5\overrightarrow{i}+6\overrightarrow{j}+4\overrightarrow{k} \right) $$

Now,

$$ \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA} $$

$$ \overrightarrow{AB}=\left( 4\overrightarrow{i}+5\overrightarrow{j}+6\overrightarrow{k} \right)-\left( 6\overrightarrow{i}+4\overrightarrow{j}+5\overrightarrow{k} \right) $$

$$ \overrightarrow{AB}=4\overrightarrow{i}+5\overrightarrow{j}+6\overrightarrow{k}-6\overrightarrow{i}-4\overrightarrow{j}-5\overrightarrow{k} $$

$$ \overrightarrow{AB}=-2\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k} $$

$$ \overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB} $$

$$ \overrightarrow{BC}=\left( 5\overrightarrow{i}+6\overrightarrow{j}+4\overrightarrow{k} \right)-\left( 4\overrightarrow{i}+5\overrightarrow{j}+6\overrightarrow{k} \right) $$

$$ \overrightarrow{BC}=5\overrightarrow{i}+6\overrightarrow{j}+4\overrightarrow{k}-4\overrightarrow{i}-5\overrightarrow{j}-6\overrightarrow{k} $$

$$ \overrightarrow{BC}=\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k} $$

$$ \overrightarrow{CA}=\overrightarrow{OA}-\overrightarrow{OC} $$

$$ \overrightarrow{CA}=\left( 6\overrightarrow{i}+4\overrightarrow{j}+5\overrightarrow{k} \right)-\left( 5\overrightarrow{i}+6\overrightarrow{j}+4\overrightarrow{k} \right) $$

$$ \overrightarrow{CA}=6\overrightarrow{i}+4\overrightarrow{j}+5\overrightarrow{k}-5\overrightarrow{i}-6\overrightarrow{j}-4\overrightarrow{k} $$

$$ \overrightarrow{CA}=\overrightarrow{i}-2\overrightarrow{j}+\overrightarrow{k} $$

Now,

$$ \left| AB \right|=\sqrt{{{\left( -2 \right)}^{2}}+{{1}^{2}}+{{1}^{2}}}=\sqrt{6} $$

$$ \left| BC \right|=\sqrt{{{1}^{2}}+{{1}^{2}}+{{\left( -2 \right)}^{2}}}=\sqrt{6} $$

$$ \left| CA \right|=\sqrt{{{1}^{2}}+{{\left( -2 \right)}^{2}}+{{1}^{2}}}=\sqrt{6} $$

Then,

$$AB=BC=CA=\sqrt{6}$$
Hence, It triangle is equilateral triangle.

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Vector Algebra Test - 41

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Vector Algebra Test - 41

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SHARING IS CARING

Question 1

2

3

4

Infinite

Solution

Question 2

$$AP$$ is opposite side of $$DM$$

$$DM$$ is parallel to $$AP$$

$$DM$$ is perpendicular to $$AP$$

None of these

Solution

Question 3

$$0$$

$$2$$

$$-1$$

$$1$$

Solution

Question 4

$$\displaystyle 3{ a }^{ \wedge }2$$

$$\displaystyle { a }^{ \wedge }2$$

$$\left( {a_1^2 + 2a_2^2 + a_3^2} \right)$$

$$\displaystyle 4{ a }^{ \wedge }2$$

Answer Required

Solution

Question 5

$$SO$$

$$2SO$$

$$OS$$

$$2OS$$

Solution

Question 6

parallel vectors which have the original vector as their resultant.

mutually perpendicular vectors which have the original vector as their resultant.

arbitrary vectors which have the original vector as their resultant.

it is not possible to resolve a vector.

Solution

Question 7

$$\;15^{\circ}$$

$$\;30^{\circ}$$

$$\;45^{\circ}$$

$$\;60^{\circ}$$

Solution

Question 8

Collinear

Forms a triangle

In different plane

None of these

Solution

Question 9

$$12$$

$$2$$

$$0$$

$$-12$$

Solution

Question 10

Right angled triangle

Equilateral triangle

Isosceles triangle

None of these

Solution

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