Vector Algebra Test

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Vector Algebra Test - 49

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Question 1
1 / -0

Let $$ABC$$ be an acute scalene triangle, and $$O$$ and $$H$$ be its circumcentre and orthocentre respectively. Further let $$N$$ be the midpoint of $$OH$$. The value of the vector sum $$\vec {NA} + \vec {NB} + \vec {NC}$$ is

A
$$\vec {0}$$ (zero vector)

B
$$\vec {HO}$$

C
$$\dfrac {1}{2}\vec {HO}$$

D
$$\dfrac {1}{2}\vec {OH}$$

Solution

$$\overrightarrow{HE}=\overrightarrow{NE}-\overrightarrow{NH}$$
$$\Rightarrow \dfrac{1}2\overrightarrow{AH}=\dfrac{1}{2}[\overrightarrow{NB}+\overrightarrow{NC}]-\dfrac{1}{2}\overrightarrow{OH}$$ ....$$ (\because HE\perp BC,AH\perp BC$$
$$\Rightarrow \overrightarrow{HE}\parallel \overrightarrow{AH}$$
$$HE=R\cos(A),AH=2R\cos(A)\\\Rightarrow \overrightarrow{HE}=\dfrac{1}{2}\overrightarrow{AH}$$
$$E$$ midpoint of $$BE$$
$$\Rightarrow \overrightarrow{NE}=\dfrac{1}{2}[\overrightarrow{NB}+\overrightarrow{NC}]$$
$$ N$$ midpoint of $$OH$$
$$\Rightarrow \overrightarrow{NH}=\dfrac{1}{2}\overrightarrow{OH} )\\\Rightarrow \overrightarrow{NH}-\overrightarrow{NA}=\overrightarrow{NB}+\overrightarrow{NC}+\overrightarrow{H)}$$
$$\Rightarrow \overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}=\overrightarrow{HO}+\dfrac{1}{2}\overrightarrow{OH}=\dfrac{1}{2}\overrightarrow{HO}$$
Question 2
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Direction angle of a vector is $$30^{\circ}$$, then find the vector.

A
$$a\hat {i} + \sqrt {3}a\hat {j}$$

B
$$\sqrt {3}a\hat {i} + b\hat {j}$$

C
$$a\hat {i} + b\hat {j}$$

D
$$\sqrt {3}a\hat {i} + \sqrt {3}b\hat {j}$$

Solution

$$\begin{matrix} Let\, \, the\, \, po{ int }\, are\, \left( { a,\, b } \right) \, \, then \\ \overrightarrow { V } =a\cos \theta \widehat { i } +b\sin \theta \widehat { j } \\ =a\cos { 30^{ \circ } } +b\sin { 30^{ \circ } } \\ =a\times \frac { { \sqrt { 3 } } }{ 2 } \widehat { i } +b\times \frac { 1 }{ 2 } \widehat { j } \\ \therefore \overrightarrow { V } =\sqrt { 3 } a\widehat { i } +b\widehat { j } \\ \end{matrix}$$
Question 3
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In the diagram, vector C is equivalent to.

A
$$\bar{D}-\bar{A}-\bar{B}$$

B
$$\bar{D}-\bar{A}+\bar{B}$$

C
$$\bar{A}+\bar{B}-\bar{D}$$

D
$$\bar{D}+\bar{A}-\bar{B}$$

Solution

C⃗ +B⃗ =D⃗ +A⃗ Diagonal of the figure.
or C⃗ =D⃗ +A⃗ −B⃗ .
Question 4
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The position vectors of A, B are a, 6 respectively. The position vector of C is $$\dfrac {5\bar{a}}{3} -\bar{b}$$. Then 3

A
C is inside the $$\Delta OAB $$

B
C is outside the $$\Delta OAB $$ but inside the angle OAB

C
C is outside the$$\Delta OAB $$ but inside the angle OBA

D
None of these

Solution

We have given that
The position vector of $$A$$ and $$B$$ are $$a$$ and $$6$$ respectively.
$$\therefore $$
Now ,
The position vector of $$C$$ is $$\dfrac{{5\overline a }}{3} - \overline b$$,
Then, the position vector at $$C$$ os inside the $$\Delta OAB$$
Hence, the option $$(A)$$ is correct.
Question 5
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If $$\bar{a},\bar{b}, \bar{c}$$ are unit vectors such that $$\bar{a}+ \bar{b}+ \bar{c}=\bar{0}$$, then $$\bar{a}.\bar{b}+ \bar{b}.\bar{c}+ \bar{c}.\bar{a}=$$

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

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

C
$$\dfrac{1}{2}$$

D
$$-\dfrac{1}{2}$$

Solution

Given,
$$\bar{a}+\bar{b}+\bar{c}=0$$
Squaring both sides,
$$\left(\bar{a}\right)^2+\left(\bar{b}\right)^2+\left(\bar{c}\right)^2+2.\bar{a}.\bar{b}+2.\bar{b}.\bar{c}+2.\bar{c}.\bar{a}=0$$
$$1+1+1+2\left(\bar{a}.\bar{b}+\bar{b}.\bar{c}+\bar{c}.\bar{a}\right)=0$$
$$\Rightarrow\left(\bar{a}.\bar{b}+\bar{b}.\bar{c}+\bar{c}.\bar{a}\right)=-\dfrac{3}{2}$$ is the correct answer.
Question 6
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If $$\vec { PR } =2\vec { i } +\vec { j } +\vec { k } ,\vec { QS } =-\vec { i } +3\vec { j } +2\vec { k } $$ then the area of the quadrilateral $$PQRS$$ is:

A
$$5\sqrt { 3 } $$

B
$$10\sqrt { 3 } $$

C
$$\cfrac { 5\sqrt { 3 } }{ 2 } $$

D
$$\cfrac { 3 }{ 2 } $$

Solution

Given $$\vec{PR}=2\vec{i}+\vec{j}+\vec{k}, \vec{QS}=-\vec{i}+3\vec{j}+2\vec{k}$$

We have to find the area of the quadrilateral $$PQRS$$

Since $$PQRS$$ is a quadrilateral, $$\vec{PR}, \vec{QS}$$ are diagonals.

Thus the area of the quadrilateral $$PQRS$$ $$=\dfrac{1}{2} |\vec{PR} \times \vec{QS}|$$

Now let us find $$\vec{PR} \times \vec{QS}$$

$$\vec{PR} \times \vec{QS}=\begin{vmatrix}\vec{i} & \vec{j} & \vec{k} \\ 2&1&1 \\ -1&3&2\end{vmatrix}$$

$$=(2-3)\vec{i}-(4+1)\vec{j}+(6+1)\vec{k}$$

$$\therefore \vec{PR} \times \vec{QS}=-\vec{i}-5\vec{j}+7\vec{k}$$

Now $$|\vec{PR} \times \vec{QS}|=\sqrt{(-1)^2+(-5)^2+7^2}=\sqrt{75}=5\sqrt{3}$$

Thus the area of the quadrilateral $$PQRS$$ $$=\dfrac{5\sqrt{3}}{2} $$
Question 7
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If $$\overline {e} = l\overline {i} + m\overline {j} + n\overline {k}$$ is a unit vector, the maximum value of $$lm + mn + nl$$ is

A
$$-\dfrac {1}{2}$$

B
$$0$$

C
$$1$$

D
$$\dfrac {3}{2}$$

Solution

Given $$\vec{e}$$ is unit vector,

$$\therefore l^2+m^2+n^2=1\dots \dots(1)$$

We know that the square of any real number is always greater than zero.

$$\Rightarrow (l+m+n)^2\ge0$$

$$\Rightarrow l^2+m^2+n^2+2(lm+mn+nl)\ge 0$$

From $$(1)$$, we get

$$lm+nm+nl\ge-\dfrac{1}{2}$$
Question 8
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If $$|\vec {a}| = 3, |\vec {b}| = 4$$ and $$|\vec {a} - \vec {b}| = 7$$ then $$|\vec {a} + \vec {b}| =$$

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

We know that
$$\Rightarrow| \vec{a}-\vec{b} |^2=| \vec{a} |^2 +| \vec{b} |^2-2| \vec{a} || \vec{b} |\cos{\theta}$$
$$\Rightarrow 7^2=3^2+4^2-2(3)(4)\cos{\theta}$$
$$\Rightarrow 49-9-16=-24\cos{\theta}$$
$$\Rightarrow \cos{\theta}=-1$$

$$\Rightarrow| \vec{a}+\vec{b} |^2=| \vec{a} |^2 +| \vec{b} |^2+2| \vec{a} || \vec{b} |\cos{\theta}$$
$$\Rightarrow| \vec{a}+\vec{b} |^2=3^2 +4^2+2(3)(4)(-1)$$
$$\Rightarrow| \vec{a}+\vec{b} |^2=9+16-24$$
$$\Rightarrow| \vec{a}+\vec{b} |^2=1$$
$$\Rightarrow| \vec{a}+\vec{b} |=1$$

Hence, answer is option $$A$$
Question 9
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The projection of the vector $$\hat {i} - 2\hat {j} + \hat {k}$$ on the vector $$4\hat {i} - 4\hat {j} + 7\hat {k}$$ is

A
$$\dfrac {5}{19}\sqrt {5}$$

B
$$\dfrac {19}{9}$$

C
$$\dfrac {9}{19}$$

D
$$\dfrac {1}{19}\sqrt {6}$$

Solution

Projection of $$\vec a$$ on $$\vec b=\dfrac {\vec a\cdot\vec b}{|\vec b|}$$

Projection of $$\hat i-2\hat j+\hat k$$ on vector $$4\hat i-4\hat j+7\hat k=\dfrac {4+8+7}{\sqrt {4^2+4^2+7^2}}=\dfrac {19}9$$

So, Option B is correct.
Question 10
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$$\overline { a } ,\overline { b } ,\overline { c } $$ are three vectors such that $$\left| \overline { a } \right| =1,\left| \overline { b } \right| =2,\left| \overline { c } \right| =3$$ and $$\overline { b } ,\overline { c } $$ are perpendicular. IF projection of $$\overline { b } $$ on $$\overline { a } $$ is the same as the projection of $$\overline { c } $$ on $$\overline { a } $$, then $$\left| \overline { a } -\overline { b } +\overline { c } \right| $$

A
$$\sqrt { 2 } $$

B
$$\sqrt { 7 } $$

C
$$\sqrt { 14 } $$

D
$$\sqrt { 21 } $$

Solution

$$\textbf{Step1-Apply condition of Dot product of two vectors}$$
$$\text{It is given that projection of }$$$$\vec{b}$$ $$\text{on}$$ $$\vec{a}$$ $$\text{is same as the projection of}$$ $$\vec{c}$$ on $$\vec{a}$$.
$$\implies \vec{b}.\vec{a} = \vec{c}.\vec{a}$$ ...[1]
$$\text{Also,}$$ $$\vec{b}$$ $$\text{and}$$ $$\vec{c}$$ $$\text{are perpendicular}$$
$$\implies \vec{b}.\vec{c} = 0$$ ...[2]
$$\text{Now,}$$
$$|\vec{a} - \vec{b} + \vec{c}|^2 = |\vec{a}|^2 +|\vec{b}|^2 +|\vec{c}|^2 -2\vec{b}.\vec{a} +2\vec{c}.\vec{a} -2\vec{b}.\vec{c} $$
$$|\vec{a} - \vec{b} + \vec{c}|^2 = |\vec{a}|^2 +|\vec{b}|^2 +|\vec{c}|^2 +0 $$ ...(Using [1] and [2])
$$|\vec{a} - \vec{b} + \vec{c}|^2 = 1+4+9 = 14$$
$$\implies |\vec{a} - \vec{b} + \vec{c}| = \sqrt{14}$$
$$\textbf{Hence , option C is correct}$$

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 49

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Vector Algebra Test - 49

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SHARING IS CARING

Question 1

$$\vec {0}$$ (zero vector)

$$\vec {HO}$$

$$\dfrac {1}{2}\vec {HO}$$

$$\dfrac {1}{2}\vec {OH}$$

Solution

Question 2

$$a\hat {i} + \sqrt {3}a\hat {j}$$

$$\sqrt {3}a\hat {i} + b\hat {j}$$

$$a\hat {i} + b\hat {j}$$

$$\sqrt {3}a\hat {i} + \sqrt {3}b\hat {j}$$

Solution

Question 3

$$\bar{D}-\bar{A}-\bar{B}$$

$$\bar{D}-\bar{A}+\bar{B}$$

$$\bar{A}+\bar{B}-\bar{D}$$

$$\bar{D}+\bar{A}-\bar{B}$$

Solution

Question 4

C is inside the $$\Delta OAB $$

C is outside the $$\Delta OAB $$ but inside the angle OAB

C is outside the$$\Delta OAB $$ but inside the angle OBA

None of these

Solution

Question 5

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

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

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

$$-\dfrac{1}{2}$$

Solution

Question 6

$$5\sqrt { 3 } $$

$$10\sqrt { 3 } $$

$$\cfrac { 5\sqrt { 3 } }{ 2 } $$

$$\cfrac { 3 }{ 2 } $$

Solution

Question 7

$$-\dfrac {1}{2}$$

$$0$$

$$1$$

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

Solution

Question 8

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 9

$$\dfrac {5}{19}\sqrt {5}$$

$$\dfrac {19}{9}$$

$$\dfrac {9}{19}$$

$$\dfrac {1}{19}\sqrt {6}$$

Solution

Question 10

$$\sqrt { 2 } $$

$$\sqrt { 7 } $$

$$\sqrt { 14 } $$

$$\sqrt { 21 } $$

Solution

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