Vector Algebra Test

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Vector Algebra Test - 52

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Question 1
1 / -0

If $$\vec{a}, \vec{b}, \vec{c}$$ are three non-coplanar vectors such that $$\vec{d}\cdot \vec{a}=\vec{d}\cdot \vec{b}=\vec{d}\cdot \vec{c}=0$$, then $$\vec{d}$$ is :-

A
a scalar

B
a null vector

C
has non-zero magnitude

D
an imaginary number

Solution

$$\because \overline { a } ,\overline { b }\ and\quad \overline { c } $$ are not coplanar vectors.
Hence,$$\overline { d } $$ can't be perpendicular to $$\overline { a } ,\overline { b }\ and\quad \overline { c } $$ at the same time.
$$ \therefore$$ The only way for $$\overline {d} .\overline{a} = 0,\ \overline { d } .\overline { b } =0$$ and $$\overline { d } .\overline { c } =0$$ to be true is if $$\overline { d } $$ is a null vector.
$$ \therefore B)Ans.$$
Question 2
1 / -0

If $$\overline {OA}=i+j+k, \overline {AB}=3i-2j+k,\overline {BC}=i+2j-2k$$ and $$\overline {CD}=2i+j+3k $$ then find the vector $$\overline{OD}$$.

A
$$7i-2j-6k$$

B
$$7i+2j+3k$$

C
$$7i+2j+5k$$

D
None of these

Solution

$$\overrightarrow {AB} = 3\hat i - 2\hat j + \hat k$$
$$ \Rightarrow \overrightarrow {OB} - \overrightarrow {OA} = 3i - 2\hat j + \hat k$$
$$ \Rightarrow \overrightarrow {OB} = 3i - 2\hat j + \hat k + \overrightarrow {OA} $$
$$ \Rightarrow \overrightarrow {OB} = 3i - 2\hat j + \hat k + \left( {\hat i + \hat j + \hat k} \right)$$
$$ \Rightarrow \overrightarrow {OB} = 4\hat i - \hat j + 2\hat k$$
Now,
$$\overrightarrow {BC} = \overrightarrow {OC} - \overrightarrow {OB} $$
$$ \Rightarrow \overrightarrow {OC} = \overrightarrow {BC} + \overrightarrow {OB} $$
$$ \Rightarrow \overrightarrow {OC} = \left( {i + 2j - 2k} \right) + \left( {4\hat i - \hat j + 2\hat k} \right)$$
$$ \Rightarrow \overrightarrow {OC} = 5\hat i + \hat j$$
similarly, $$\overrightarrow {CD} = \overrightarrow {OD} - \overrightarrow {OC} $$
$$ \Rightarrow \overrightarrow {OD} = \overrightarrow {CD} + \overrightarrow {OC} $$
$$ \Rightarrow \overrightarrow {OD} = \left( {2\hat i + \hat j + 3\hat k} \right) + \left( {5\hat i + \hat j} \right)$$
$$ \Rightarrow \overrightarrow {OD} = 7\hat i + 2\hat j + 3\hat k$$
Question 3
1 / -0

The value of $$\bar {a} \times (\bar {b}+\bar {c})+\bar {b}\times (\bar {c}+\bar {a})+\bar {c}\times (\bar {a}+\bar {b})=$$

A
$$0$$

B
$$-1$$

C
$$2\ \bar {a}\times (\bar {b}+\bar {c})$$

D
$$[\bar {a}\bar {b}\bar {c}]$$

Solution

$$\bar {a} \times (\bar{b} + \bar{c}) + \bar{b} \times (\bar{c} + \bar{a}) + \bar{c} \times (\bar{a} + \bar{b})$$

$$ = \bar {a} \times \bar{b} + \bar{a} \times \bar{c} + \bar{b} \times \bar{c} + \bar{b} \times \bar{a} + \bar{c} \times \bar{a} + \bar{c} \times \bar{b}$$

$$ = \bar {a} \times \bar{b} + \bar{a} \times \bar{c} + \bar{b} \times \bar{c} - \bar{a} \times \bar{b} - \bar{a} \times \bar{c} - \bar{b} \times \bar{c}$$

$$ = 0 $$(Option A)

Formula:
$$\bar {b} \times \bar {a} = - \bar{a} \times \bar{b}$$
Question 4
1 / -0

Let $$\vec{a},\vec{b}$$ and $$\vec{c}$$ be three non-zero vectors such that no two of these are collinear. If the vector $$\vec{a}+2\vec{b}$$ is collinear with $$\vec{c}$$ and $$\vec{b}+3\vec{c}$$ is collinear with $$\vec{a}(\lambda$$ being some non-zero scalar), then $$\vec{a}+2\vec{ b}+6\vec{c}$$ equals

A
$$\lambda\vec{a}$$

B
$$\lambda\vec{b}$$

C
$$\lambda\vec{c}$$

D
$$\vec{0}$$

Solution

$$\vec { a } +2\vec { b } =\lambda\vec { c } \quad -(i)$$
$$ \vec { b } +3\vec { c } =\mu\vec { a } \quad -(ii)$$
$$ Substituting\quad \vec { a } \quad from\quad (i)\quad in\quad (ii)$$
$$\implies\quad \vec { b } +3\vec { c } =\mu(\lambda\vec { c } -2\vec { b } )$$
$$\implies\quad (1+2\mu)\vec { b } =(\lambda\mu-3)\vec { c } $$
$$\implies\quad But\quad since\quad \vec { b } \quad and\quad \vec { c } \quad are\quad not\quad collinear.$$
$$\implies\quad Either\quad 1+2\mu=0,\quad \lambda\mu-3=0$$
$$\implies\quad \mu=\cfrac { -1 }{ 2 } $$
$$ and\quad \cfrac { -\lambda}{ 2 } =3$$
$$\implies\quad \lambda=-6$$

$$ Substituting\quad in\quad (i)$$
$$\implies\quad \vec { a } +2\vec { b } +6\vec { c } =\vec { 0 } $$

$$ \therefore D)Answer.$$
Question 5
1 / -0

The $$P.V.'s$$ of the vertices of a $$\triangle ABC$$ are $$\bar {i}+\bar {j}+\bar {k}, 4\bar {i}+\bar {j}+\bar {k}, 4\bar {i}+5\bar {j}+\bar {k}$$. The $$P.V.$$ of the circumcentre of $$\triangle ABC$$ is

A
$$\dfrac{5}{2}\bar {i}+3\bar {j}+\bar {k}$$

B
$$5\bar {i}+\dfrac{3}{2}\bar {j}+\bar {k}$$

C
$$5\bar {i}+3\bar {j}+\dfrac{1}{2}\bar {k}$$

D
$$\bar {i}+\bar {j}+\bar {k}$$

Solution

We have,
$$A(1,1,1)$$ $$B(4,1,1)$$ $$C(4,5,1)$$
$$AB=3$$ $$BC=4$$ $$CA=5$$
Hence,
Triangle is right angle at $$B$$.
$$\therefore $$ CIrcumcentre is mid point of $$AC = \left( {\frac{5}{2},3,1} \right)$$
$$ = \left( {\frac{5}{2}\overline i ,3\overline j ,\overline k } \right)$$
Then,
Option $$A$$ is correct answer.
Question 6
1 / -0

Component of $$\vec{a}=\hat{i}-\hat{j}-\hat{k}$$ perpendicular to the vector $$\vec{b}=2\hat{i}+\hat{j}-\hat{k}$$ is?

A
$$\dfrac{1}{3}(\hat{i}+2\hat{j}+2\hat{k})$$

B
$$\dfrac{1}{3}(\hat{i}-4\hat{j}-2\hat{k})$$

C
$$\dfrac{1}{3}(\hat{i}+4\hat{j}+2\hat{k})$$

D
$$\dfrac{1}{3}(\hat{i}+2\hat{j}+\hat{k})$$

Solution

$$\bar{a}=\hat{i}-\hat{j}-\hat{k}$$
$$\bar{b}=2\hat{i}+\hat{j}-\hat{k}$$
$$a_{\bot}=a-\dfrac{a.b}{b.b}(\bar{b})$$
$$a_{\bot}=(\hat{i}-\hat{j}-\hat{k})-\dfrac{(\hat{i}-\hat{j}-\hat{k}).(2\hat{i}+\hat{j}-\hat{k})}{(\sqrt{(2)^{2}+(1)^{2}+(1)^{2}})^{2}} (2\hat{i}+\hat{j}-\hat{k})$$
$$a_{\bot}=(\hat{i}-\hat{j}-\hat{k})-\dfrac{(2-1+1)}{6}(2\hat{i}+\hat{j}-\hat{k})$$
$$a_{\bot}=(\hat{i}-\hat{j}-\hat{k})-\dfrac{(2\hat{i}+\hat{j}-\hat{k})}{3}$$
$$a_{\bot}=\dfrac{(3\hat{i}-2\hat{i})-(3\hat{j}+\hat{j})-(3\hat{k}-\hat{k})}{3}$$
$$a_{\bot}=\dfrac{\hat{i}-4\hat{j}-2\hat{k}}{3}=\dfrac{1}{3}(\hat{i}-4\hat{j}-2\hat{k})$$
Question 7
1 / -0

If $$a^b=b^c=ab$$, then $$b+c$$ always equals?

A
$$\dfrac{1}{bc}$$

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

C
$$1$$

D
$$bc$$

Solution

$${ a }^{ b }={ b }^{ c }=ab=p(let)$$
$$\implies\quad { a }^{ b }=p$$
taking log
$$\implies\quad b\log { a } =\log { p } \quad -(i)$$
$$ { b }^{ c }=p$$
Taking log
$$\implies\quad c\log { b } =\log { p } \quad -(ii)$$
$$ ab=p$$
Taking log
$$ \log { a } +\log { b } =\log { p } $$
Using$$ (i)\& (ii),$$we can say:
$$ \cfrac { \log { p } }{ b } +\cfrac { \log { p } }{ c } =\log { p } $$
$$\implies\quad \cfrac { 1 }{ b } +\cfrac { 1 }{ c } =1$$
$$\implies\quad b+c=bc$$
Question 8
1 / -0

$$\overline a = \overline i - \overline k ,\overline b = x\overline i + \overline j + (1 - x)\overline k $$ and $$\overline c = y\overline i + \lambda \overline j + (1 + x - y)\overline k $$ then $$\left[ {\overline a \overline b \overline c } \right]$$ depends on:-

A
neither x nor y

B
both x and y

C
only x

D
only y

Solution

$$\textbf{Step1-Find scalar triple product of vcectors}$$
$$\overline { a } .[\overline { b } \times \overline { c } ]$$
$$ \left| \begin{matrix} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & \lambda & 1+x-y \end{matrix} \right| $$
$$\implies\quad 1(1+x-y-\lambda +\lambda x)-1(\lambda x-y)$$
$$\implies\quad 1+x-y-\lambda +\lambda x-\lambda x+y$$
$$ \therefore [\begin{matrix} a & b & c \end{matrix}]$$ $$\text{depends only on x,not y.}$$
$$\textbf{Hence , option C is correct}$$
Question 9
1 / -0

$$A(1, -1, -3)$$, $$B(2, 1, -2)$$ & $$C(-5, 2, -6)$$ are the position vectors of the vertices of a triangle ABC. The length of the bisector of its internal angle at A is?

A
$$\sqrt{10}/4$$

B
$$3\sqrt{10}/4$$

C
$$\sqrt{10}$$

D
None

Solution

$$AB=\sqrt { { (2-1) }^{ 2 }+{ (1+1) }^{ 2 }+{ (-2+3) }^{ 2 } } $$
$$ =\sqrt { 1+4+1 } $$
$$ =\sqrt { 6 } $$
$$ AC=\sqrt { { (-5-1) }^{ 2 }+{ (2+1) }^{ 2 }+{ (-6+3) }^{ 2 } } $$
$$ =\sqrt { 36+9+9 } $$
$$=\sqrt { 54 } $$
$$ =3\sqrt { 6 } $$
$$ \cfrac { AB }{ AC } =\cfrac { BE }{ EC } (\because \triangle AEB\sim \triangle AEC\quad \therefore \angle BAE=\angle EAC)$$
$$\implies\quad \cfrac { BE }{ EC } =\cfrac { AB }{ AC } =\cfrac { \sqrt { 6 } }{ 3\sqrt { 6 } } =\cfrac { 1 }{ 3 } $$
Using section formula,
$$ E=(\cfrac { -5+2(3) }{ 1+3 } ,\cfrac { 2+3(1) }{ 1+3 } ,\cfrac { -6+3(-2) }{ 4 } )$$
$$\implies\quad E=(\cfrac { -5+6 }{ 4 } ,\cfrac { 2+3 }{ 4 } ,\cfrac { -6-6 }{ 4 } )$$
$$ =(\cfrac { 1 }{ 4 } ,\cfrac { 5 }{ 4 } ,{ -3 })$$
$$ \therefore Length\quad AE=\sqrt { { (1-\cfrac { 1 }{ 4 } ) }^{ 2 }+{ (-1-\cfrac { 5 }{ 4 } ) }^{ 2 }+{ (-3+3) }^{ 2 } } $$
$$ =\sqrt { \cfrac { 9 }{ 16 } +\cfrac { 81 }{ 16 } +0 } $$
$$ =\sqrt { \cfrac { 90 }{ 16 } } $$
$$ =\sqrt { \cfrac { 9\times 10 }{ 16 } } $$
$$ =\cfrac { 3\sqrt { 10 } }{ 4 } $$
Question 10
1 / -0

Let P, Q, R and S be the points on the plane with position vectors $$-2\hat{i}-\hat{j}, 4\hat{i}, 3\hat{i}+3\hat{j}$$ and $$-3\hat{i}+2\hat{j}$$ respectively. the quadrilateral PQRS must be a.

A
Parallelogram, which is neither a rhombus nor a rectangle

B
Square

C
Rectangle, but not a square

D
Rhombus, but not a square

Solution

Consider the problem
As we given
$$\left( { - 2\hat i - \hat j} \right),\left( {4\hat i} \right),\left( {3\hat i + 3\hat j} \right),\left( { - 3i + 2\hat j} \right)$$
On converting above vectors into Cartesian coordinates
Then , consider,
$$P\left( { - 2, - 1} \right),Q\left( {4,0} \right),R\left( {3,3} \right),S\left( { - 3,2} \right)$$
By distance formula

$$=\sqrt{(x_1-x_2)^2+(y_1+y_2)^2}$$

$$d(PQ)=\sqrt{(4+2)^2+(0+1)^2}=\sqrt{37}$$
Similarly,

$$d(QR)=\sqrt{10}$$

$$d(RS)={\sqrt{37}}$$

$$d(SP)=\sqrt{10}$$

Therefore,
$$d(PQ)=d(RS)$$ and $$d(QR)=d(SP)$$
Hence,
$$PQRS$$ is parallelogram
Now,
Slope $$=\frac{y_2-y_1}{x_2-x_1}$$
Therefore, let slope is $$S$$
then ,
$$S(PQ)=\frac{0-(-1)}{4-(-2)}=\frac{0+1}{4+2}=\frac{1}{6}$$
Similarly,
$$S(QR)=\frac{3}{-1}=-3$$

$$S(RS)=\frac{-1}{-6}=\frac{1}{6}$$

$$S(PS)=-3$$

$$S(PQ)=S(RS)$$ this implies $$PQ \parallel RS$$
And
$$S(QR)=S(SP)$$ this implies $$QR \parallel SP$$

$$S(PQ) \ne S(QR)$$ i.e. they aren't parallel

$$S(PQ) \times S(QR) \ne -1$$ i.e. they aren't perpendicular.

Therefore

$$PQRS$$ is a parallelogram but it is nor a rhombus and not rectangle.

Hence option $$A$$ is the correct answer.

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Vector Algebra Test - 52

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Vector Algebra Test - 52

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SHARING IS CARING

Question 1

a scalar

a null vector

has non-zero magnitude

an imaginary number

Solution

Question 2

$$7i-2j-6k$$

$$7i+2j+3k$$

$$7i+2j+5k$$

None of these

Solution

Question 3

$$0$$

$$-1$$

$$2\ \bar {a}\times (\bar {b}+\bar {c})$$

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

Solution

Question 4

$$\lambda\vec{a}$$

$$\lambda\vec{b}$$

$$\lambda\vec{c}$$

$$\vec{0}$$

Solution

Question 5

$$\dfrac{5}{2}\bar {i}+3\bar {j}+\bar {k}$$

$$5\bar {i}+\dfrac{3}{2}\bar {j}+\bar {k}$$

$$5\bar {i}+3\bar {j}+\dfrac{1}{2}\bar {k}$$

$$\bar {i}+\bar {j}+\bar {k}$$

Solution

Question 6

$$\dfrac{1}{3}(\hat{i}+2\hat{j}+2\hat{k})$$

$$\dfrac{1}{3}(\hat{i}-4\hat{j}-2\hat{k})$$

$$\dfrac{1}{3}(\hat{i}+4\hat{j}+2\hat{k})$$

$$\dfrac{1}{3}(\hat{i}+2\hat{j}+\hat{k})$$

Solution

Question 7

$$\dfrac{1}{bc}$$

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

$$1$$

$$bc$$

Solution

Question 8

neither x nor y

both x and y

only x

only y

Solution

Question 9

$$\sqrt{10}/4$$

$$3\sqrt{10}/4$$

$$\sqrt{10}$$

None

Solution

Question 10

Parallelogram, which is neither a rhombus nor a rectangle

Square

Rectangle, but not a square

Rhombus, but not a square

Solution

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