Vector Algebra Test

Result

Vector Algebra Test - 42

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If a and b are two non-zero and non-collinear vectors, then a + b and a - b are

A
Linearly dependent vectors

B
Linearly dependent and independent vectors

C
Linearly independent vectors

D
None of these

Solution
Question 2
1 / -0

Let P, Q, R and S be the points on the plane with position vectors (-2i - j), 4i, (3i + 3j) and (-3i + 2j) respectively. The quadilateral 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 $$=\dfrac{y_2-y_1}{x_2-x_1}$$
Therefore, let slope is $$S$$
then ,
$$S(PQ)=\dfrac{0-(-1)}{4-(-2)}=\dfrac{0+1}{4+2}=\dfrac{1}{6}$$
Similarly,
$$S(QR)=\dfrac{3}{-1}=-3$$

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

$$S(PS)=-3$$

$$S(PQ)=S(RS)$$ 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.
Question 3
1 / -0

In triangle $$ABC$$, which of the following is not true?

A
$$\vec {AB}+\vec {BC}+\vec {CA}=\vec {0}$$

B
$$\vec {AB}+\vec {BC}-\vec {AC}=\vec {0}$$

C
$$\vec {AB}+\vec {BC}-\vec {CA}=\vec {0}$$

D
$$\vec {AB}-\vec {CB}+\vec {CA}=\vec {0}$$

Solution

$${\textbf{Step 1: Using triangle law of vector addition, find the resultant of two vectors}}$$
$$ \Rightarrow \overrightarrow {AB}  + \overrightarrow {BC}  = \overrightarrow {AC} $$
$${\textbf{Step 2: Solve the vector sum and Compare with the given options}}$$
$$ \Rightarrow \overrightarrow {AB}  + \overrightarrow {BC}  - \overrightarrow {AC}  = \overrightarrow 0 $$
$${\text{Or }}\overrightarrow {AB}  + \overrightarrow {BC}  + \overrightarrow {CA}  = \overrightarrow 0 $$
$${\text{Or }}\overrightarrow {AB}  + \overrightarrow {BC}  + \overrightarrow {CA}  = \overrightarrow 0 $$
$${\text{So, }}$$$$\overrightarrow {AB}  + \overrightarrow {BC}  - \overrightarrow {CA}  \ne \overrightarrow 0 $$
$${\textbf{Hence, Option (C) is incorrect}}$$
Question 4
1 / -0

$$\vec{a}\, \neq\, \bar{0},\,\vec{b}\,\neq\,\bar{0},\,\vec{c}\,\neq\,0,\,\vec{a}\,\times\,\vec{b}\,=\,\vec{0},\, \bar{b}\,\times\,\vec{c}\,=\,0\,\Rightarrow\,\vec{a}\,\times\,\vec{c}\,=$$

A
$$\displaystyle\vec{b}$$

B
$$\displaystyle\vec{a}$$

C
$$\displaystyle\vec{0}$$

D
$$\displaystyle\vec{i}\,+\,\vec{j}\,+\,\vec{k}$$

Solution

We know,
$$\overrightarrow{a}\times \overrightarrow{b}=0$$ $$\Rightarrow \overrightarrow{a}=\lambda\overrightarrow{b}$$
And since $$\overrightarrow{b}\times \overrightarrow{c}=0$$ $$\overrightarrow{c}=\lambda '\overrightarrow{b} $$
So, $$\overrightarrow{a}\times \overrightarrow{c}=\lambda\times\lambda '\times \overrightarrow{b}\times \overrightarrow{b}=0$$
So, Option (C)
Question 5
1 / -0

The position vectors of $$P$$ and $$Q$$ are respectively $$a$$ and $$b$$. If $$R$$ is a point on $$PQ$$, $$PQ$$ such that $$PR=5PQ$$, then the position vector of $$R$$ is

A
$$5b-4a$$

B
$$5b+4a$$

C
$$4b-5a$$

D
$$4b+5a$$

Solution

Given condition
$$PR=5PQ$$
$$R-P=5(Q-P)$$
$$R=5Q-5P+P$$
$$R=5Q-4P$$
$$R=5b-4a$$
Question 6
1 / -0

If $$|\vec {a}| = |\vec {b}| = 1$$ and $$|\vec {a} + \vec {b}| = \sqrt {3}$$, then the value of $$(3\vec {a} - 4\vec {b}) \cdot (2\vec {a} + 5\vec {b})$$ is

A
$$-21$$

B
$$-\dfrac {21}{2}$$

C
$$21$$

D
$$\dfrac {21}{2}$$

Solution

$$\because |\vec {a} + \vec {b}| = \sqrt {3}$$
$$\Rightarrow |\vec {a}|^{2} + |\vec {b}|^{2} + 2\vec {a} \cdot \vec {b} = 3$$
$$\Rightarrow \vec {a}\cdot \vec {b} = \dfrac {1}{2}$$ ..... (i)
$$(\because |\vec {a}| = |\vec {b}| = 1 given)$$
$$\therefore (3\vec {a} - 4\vec {b})\cdot (2\vec {a} + 5\vec {b})$$
$$= 6|\vec {a}|^{2} + 15\vec {a}\cdot \vec {b} - 8\vec {a}\cdot \vec {b} - 20 |\vec {b}|^{2}$$
$$= 6 + 7\vec {a}\cdot \vec {b} - 20$$
$$= 6 + \dfrac {7}{2} - 20$$ [from Eq. (i)]
$$= \dfrac {7 - 28}{2} = -\dfrac {21}{2}$$
Question 7
1 / -0

Let $$\vec {a} = \vec {i} + 2\vec {j} + \vec {k}, \vec {b} = \vec {i} - \vec {j} + \vec {k}$$ and $$\vec {c} = \vec {i} + \vec {j} - \vec {k}$$. A vector in the plane of $$\vec {a}$$ and $$\vec {b}$$ has projection $$\dfrac {1}{\sqrt {3}} \ on\ \vec {c}$$. Then, one such vector is

A
$$4\vec {i} + \vec {j} - 4\vec {k}$$

B
$$3\vec {i} + \vec {j} - 3\vec {k}$$

C
$$4\vec {i} - \vec {j} + 4\vec {k}$$

D
$$2\vec {i} + \vec {j} - 2\vec {k}$$

Solution

Let $$\vec { v } $$ be the vector then $$\vec { v } =\vec { a } +\lambda \vec { b } $$
$$\therefore \vec { v } =\left( 1+\lambda \right) \hat { i } +\left( 2-\lambda \right) \hat { j } +\left( 1+\lambda \right) \hat { k } $$

Now $$\left| \dfrac { \vec { v } \cdot \vec {c } }{ \left| \vec { c } \right| } \right| =\dfrac { 1 }{ \sqrt { 3 } } $$
$$\Rightarrow \dfrac { \left| 1+\lambda +2-\lambda -1-\lambda \right| }{ \sqrt { 3 } } =\dfrac { 1 }{ \sqrt { 3 } } $$
$$\Rightarrow 2-\lambda =\pm 1$$

$$\Rightarrow \lambda =3$$ or $$1$$

$$\therefore \vec { v } =4\hat { i } -\hat { j } +4\hat { k } $$ or $$2\hat { i } +\hat { j } +2\hat { k } $$
Question 8
1 / -0

Let $$\overrightarrow {a} , \overrightarrow {b}$$ and $$\overrightarrow {c}$$ be vectors with magnitudes 3, 4 and 5 respectively and $$\overrightarrow{a} + \overrightarrow {b}+\overrightarrow {c}=\overrightarrow {0}$$, then the value of $$\overrightarrow{a}. \overrightarrow{b}+\overrightarrow{b}. \overrightarrow{c} + \overrightarrow{c}. \overrightarrow{a}$$ is

A
47

B
25

C
50

D
-25

Solution

$$ \vec{a}+\vec{b}+\vec{c} = \vec{0}. $$
dot product with $$ \vec{a},\vec{b} $$ and $$ \vec{c} $$ both sides
$$ \vec{a}.(\vec{a}+\vec{b}+\vec{c}) = \vec{0}\Rightarrow (\vec{a})^{2}+\vec{a}.\vec{b}+\vec{a}.\vec{c} = \vec{0}. $$
Similarly, $$ \vec{b},\vec{a}+(\vec{b})^{2}+(\vec{b}.\vec{c}) = \vec{0}; \vec{c}.\vec{a}+\vec{b}.\vec{c}+(\vec{c})^{2} = \vec{0} $$
adding all 3 equation,
$$ (\vec{a})^{2}+(\vec{b})^{2}+(\vec{c})^{2}+2(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}) = \vec{0} $$
$$ 2(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}) = -[3^{2}+4^{2}+5^{2}] = -50 $$
$$ \boxed { \vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a} = -25} $$
Question 9
1 / -0

Find the correct vectorial relationship with the help of the figure above.

A
$$\vec {x} + \vec {y} = \vec {z}$$

B
$$\vec {y} + \vec {z} = \vec {x}$$

C
$$\vec {x} + \vec {z} = \vec {y}$$

D
$$\vec {z} - \vec {x} = \vec {y}$$

E
$$\vec {z} - \vec {y} = \vec {x}$$

Solution

If we observe clearly, we can see that $$y$$ is resultant of $$x$$ and $$z$$.
So, we get $$y = x+z$$.
Question 10
1 / -0

How much does a watch lose per day, if its hands coincide every $$64$$ minutes?

A
$$32\cfrac { 8 }{ 11 } min.$$

B
$$31\cfrac { 8 }{ 11 } min.$$

C
$$32\cfrac { 3 }{ 11 } min.$$

D
$$None\ of\ these$$

Solution

$$55min.$$ spaces are covered in $$60min.$$

$$60min.$$ spaces are covered in $$\left( \cfrac { 60 }{ 55 } \times 60 \right) min=65\cfrac { 5 }{ 11 } min.$$

Loss in $$64min.=\left( 65\cfrac { 5 }{ 11 } -64 \right) =\cfrac { 16 }{ 11 } min.$$

Loss in $$24hrs.=\left( \cfrac { 16 }{ 11 } \times \cfrac { 1 }{ 64 } \times 24\times 60 \right) min.=32\cfrac { 8 }{ 11 } min$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 42

Result

Vector Algebra Test - 42

Score

-

Rank

-

SHARING IS CARING

Question 1

Linearly dependent vectors

Linearly dependent and independent vectors

Linearly independent vectors

None of these

Solution

Question 2

Parallelogram, which is neither a rhombus nor a rectangle

Square

Rectangle, but not a square

Rhombus, but not a square

Solution

Question 3

$$\vec {AB}+\vec {BC}+\vec {CA}=\vec {0}$$

$$\vec {AB}+\vec {BC}-\vec {AC}=\vec {0}$$

$$\vec {AB}+\vec {BC}-\vec {CA}=\vec {0}$$

$$\vec {AB}-\vec {CB}+\vec {CA}=\vec {0}$$

Solution

Question 4

$$\displaystyle\vec{b}$$

$$\displaystyle\vec{a}$$

$$\displaystyle\vec{0}$$

$$\displaystyle\vec{i}\,+\,\vec{j}\,+\,\vec{k}$$

Solution

Question 5

$$5b-4a$$

$$5b+4a$$

$$4b-5a$$

$$4b+5a$$

Solution

Question 6

$$-21$$

$$-\dfrac {21}{2}$$

$$21$$

$$\dfrac {21}{2}$$

Solution

Question 7

$$4\vec {i} + \vec {j} - 4\vec {k}$$

$$3\vec {i} + \vec {j} - 3\vec {k}$$

$$4\vec {i} - \vec {j} + 4\vec {k}$$

$$2\vec {i} + \vec {j} - 2\vec {k}$$

Solution

Question 8

47

25

50

-25

Solution

Question 9

$$\vec {x} + \vec {y} = \vec {z}$$

$$\vec {y} + \vec {z} = \vec {x}$$

$$\vec {x} + \vec {z} = \vec {y}$$

$$\vec {z} - \vec {x} = \vec {y}$$

$$\vec {z} - \vec {y} = \vec {x}$$

Solution

Question 10

$$32\cfrac { 8 }{ 11 } min.$$

$$31\cfrac { 8 }{ 11 } min.$$

$$32\cfrac { 3 }{ 11 } min.$$

$$None\ of\ these$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.