Vector Algebra Test

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Vector Algebra Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

The unit vector in the direction of $$\overrightarrow{a}$$ is

A
$$\dfrac{\vec{a}}{|\vec{a}|}$$

B
$$\vec{a}|\vec{a}|$$

C
$$a^2$$

D
$$\hat{i}$$

Solution

Consider the given vector $$\vec a$$.

Unit vector $$\hat a=\dfrac{\vec a}{|\vec a|}$$

Hence, this is the answer.
Question 2
1 / -0

If $$\vec{a}$$ be the position vector whose tip is (5,-3), find the coordinates of a point B such that $$ \vec{AB} = \vec{a},$$ the coordinates of A being (4,-1).

A
(9, -4)

B
(-9, -4)

C
(9, 4)

D
none of these

Solution

$$\vec{a}$$ is position vector of $$(5, -3)$$
$$\vec{a}=5\hat{i}-3\hat{j}$$
Coordinates of $$A=(4, -1)$$
$$\vec{OA}=4\hat{i}-\hat{j}$$
$$\vec{AB}=\vec{a}=5\hat{i}-3\hat{j}$$
$$\vec{OB}-\vec{OA}=5\hat{i}-3\hat{j}$$
$$\Rightarrow \vec{OB}=5\hat{i}-3\hat{j}+\vec{OA}$$
$$=5\hat{i}-3\hat{j}+4\hat{i}-3\hat{j}$$
$$=a\hat{i}-4\hat{j}$$
$$\vec{OB}=9\hat{i}-4\hat{j}$$
Coordinates of point B$$=(9, -4)$$.
Question 3
1 / -0

find the coordinate of the tip of the position vector which is equivalent to $$ \vec{AB}$$, where the coordinates of A and B are (-1, 3) and (-2, 1) respectively.

A
(+1,+2)

B
(+1,-2)

C
(-1,+2)

D
(-1,-2)

Solution

$$A(-1, 3)$$
$$B(-2, 1)$$
'O' be the origin
$$\vec{OA}=-\hat{i}+3\hat{j}$$; $$\vec{OB}=-2\hat{i}+\hat{j}$$
$$\vec{AB}=\vec{OB}-\vec{OA}=(-2\hat{i}+\hat{j})-(-\hat{i}+3\hat{j})$$
$$=-2\hat{i}+\hat{j}+\hat{i}-3\hat{j}$$
$$=-\hat{i}-2\hat{j}$$
$$\therefore$$ Coordinates $$=(-1, -2)$$.
Question 4
1 / -0

If the position vectors of the points $$A(3,4),B(5, -6)$$ and $$C(4,-1)$$ are $$ \vec{a}, \vec{b}, \vec{c}$$ respectively, compute $$ \vec{a}+2\vec{b}-3\vec{c}. $$

A
$$ -5\hat{i}-1\hat{j} $$

B
$$ \hat{i}-5\hat{j} $$

C
$$ \hat{i}+5\hat{j} $$

D
none of these

Solution

Position vectors of points
$$A(3, 4), B(5, -6), C(-4, -1)$$ are $$\vec{a}, \vec{b}, \vec{c}$$
$$\vec{a}=3\hat{i}+4\hat{j}$$
$$\vec{b}=5\hat{i}-6\hat{j}$$
$$\vec{c}=4\hat{i}-\hat{j}$$
$$\vec{a}+2\vec{b}-3\vec{c}=3\hat{i}+4\hat{j}+2(5\hat{i}-6\hat{j})-3(4\hat{i}-\hat{j})$$
$$=3\hat{i}+4\hat{j}+10\hat{i}-12\hat{j}-12\hat{i}+3\hat{j}$$
$$=\hat{i}-5\hat{j}$$
$$\vec{a}+2\vec{b}-3\vec{c}=\hat{i}-5\hat{j}$$.
Question 5
1 / -0

If $$\vec a$$ is parallel to $$\vec b \times \vec c$$, then $$(\vec a \times \vec b) \cdot (\vec a \times \vec c)$$ is equal to

A
$$\mid \vec a \mid^2 (\vec b \cdot \vec c)$$

B
$$\mid \vec b \mid^2 (\vec a \cdot \vec c)$$

C
$$\mid \vec c \mid^2 (\vec a \cdot \vec b)$$

D
none of these

Solution

$$(\vec a \times \vec b) \cdot (\vec a \times \vec c) = (\vec a \times \vec b) \cdot \vec u = \vec a \times \vec c$$
$$\Rightarrow \vec a \cdot (\vec b \times \vec u) = \vec a \cdot [\vec b \times (\vec a \times \vec c)]$$
$$ = \vec a \cdot [(\vec b \cdot \vec c) \vec a - (\vec a \cdot \vec b) \vec c]$$
$$ = \vec a \cdot (\vec b \cdot \vec c) \vec a $$ $$(\because \vec a \cdot \vec b = 0)$$
$$ = \mid \vec a\mid^2 (\vec b \cdot \vec c)$$
Question 6
1 / -0

If $$\mid \vec a \mid = 2$$ and $$\mid \vec b \mid = 3$$ and $$\vec a \cdot \vec d = 0$$, then $$(\vec a \times (\vec a \times (\vec a \times (\vec a \times \vec b ))))$$ is equal to

A
$$48 \hat b$$

B
$$- 48 \hat b$$

C
$$48 \hat a $$

D
$$48 \hat a$$

Solution

$$(\vec a \times (\vec a \times (\vec a \times (\vec a \times \vec b )))) = (\vec a \times ( \vec a \times ((\vec a \cdot \vec b) \vec a - (\vec a \cdot \vec b) \vec b)))$$
$$ = (\vec a \times (\vec a \times (- 4 \vec b)))$$
$$ = -4 (\vec a \times(\vec a \times \vec b))$$
$$ = -4((\vec a \cdot \vec b) \vec a - (\vec a \cdot \vec a)\vec b)$$
$$ = -4(-4 \vec b) = 16 \vec b = 48 \hat b$$
Question 7
1 / -0

Let $$\mathrm{A}\mathrm{B}\mathrm{C}$$ be a triangle and let $$\mathrm{D},\mathrm{E}$$ be the midpoints of the sides $$\mathrm{A}\mathrm{B},\mathrm{A}\mathrm{C}$$ respectively,then $$\hat { BE } +\hat { DC } =$$

A
$$\hat { BC } $$

B
$${ \dfrac { 1 }{ 2 } }\hat { BC } $$

C
$$ { \dfrac { 3 }{ 2 } \hat { BC } } $$

D
$$\dfrac { 3 }{ 4 } \hat { BC } $$

Solution

Since, $$D$$ and $$E$$ are the mid points of the sides $$AB$$ and $$AC$$
Therefore, $$OD=\dfrac{OA+OB}{2}$$ and $$OE=\dfrac{OA+OC}{2}$$
$$Now, BE+DC\\=OE-OB+OC-OD=OC-OB+\dfrac{OC-OB}{2}\\=\dfrac{3(OC-OB)}{2}=\dfrac{3BC}{2}$$
Question 8
1 / -0

The position vectors of $$A,B,C$$ are $$\overline{i}+\overline{j}+\overline{k},\ 4\overline{i}+\overline{j}+\overline{k},\ 4\overline{i}+5\overline{j}+\overline{k}$$ . Then the position vector of the circumcentre of the triangle $$ABC$$ is

A
$$3\hat{i}+2\hat{j}+\hat{k}$$

B
$$\displaystyle \frac{1}{2}(6\hat{i}+\hat{j}+\hat{k})$$

C
$$\displaystyle \frac{1}{2}(5\hat{i}+6\hat{j}+2\hat{k})$$

D
$${\dfrac{1}{2}(9\overline{i}}+7\overline{j}+3\overline{k})$$

Solution

Position vector of $$A$$ is $$\hat i+\hat j+\hat k$$, gives coordinates as $$(1,1,1)$$.
Position vector of $$B$$ is $$4\hat i+\hat j+\hat k$$, gives coordinates as $$(4,1,1)$$.
Position vector of $$C$$ is $$4\hat i+5\hat j+\hat k$$, gives coordinates as $$(4,5,1)$$.

$$\vec{AB} = \vec B - \vec A = 3\hat i$$
$$\vec {BC} = \vec C - \vec B = 4\hat j$$
$$\vec {AC} = \vec C - \vec A = 3\hat i+4\hat j$$
Here, we can see that $$\triangle ABC$$ is a right angled triangle with right angle at $$B$$.

Hence, circumcenter will be the midpoint of $$AC$$.
Mid point of $$AC = D\equiv\left(\dfrac{1+4}2 , \dfrac{1+5}2, \dfrac{1+1}2\right) = \left(\dfrac52,3,1\right)$$
So, position vector of $$D$$ will be
$$\dfrac12(5\hat i+6\hat j+2\hat k)$$
which is a circumcenter.
Question 9
1 / -0

If $$\mathrm{O}$$ is the circumcentre and $$\mathrm{O}^{'}$$ is the orthocentre of a triangle $$\mathrm{A}\mathrm{B}\mathrm{C}$$ and if $$\mathrm{A}\mathrm{P}$$ is the circumdiameter then
$$\vec{\mathrm{A}\mathrm{O}}+\vec{\mathrm{O}^{'}\mathrm{B}}+\vec{\mathrm{O}^{'}\mathrm{C}}=$$

A
$$\vec{\mathrm{O}\mathrm{A}}$$

B
$$\vec{\mathrm{O}^{'}\mathrm{A}}$$

C
$$\vec{\mathrm{A}\mathrm{P}}$$

D
$$\vec{\mathrm{A}\mathrm{O}}$$

Solution

$$\left | \vec{AP} \right |=2\left | AO \right |$$
$$=O'B+O'C+O'A-O'A+AO'$$
$$=(\bar{A}+\bar{B}+\bar{C}-3 \ \bar{O'})+2\ \overline{AO'}$$
So use $$3\ \overline{OS}=2\ \overline{OC}$$
where $$S$$ is centroid
$$C$$ is circumcentre
$$O$$ is orthocenter
$$=3(\dfrac{\bar{A}+\bar{B}+\bar{C}-\bar{O'}}{3})+2\overline{AO'}$$
$$=2\ \overline{O'O}+2\ \overline{AO'}$$
$$=2\ \overline{AO}$$
$$=\overline{AP}$$
Question 10
1 / -0

If the vectors $$\overline{a}=3\overline{i}+\overline{j}-2\overline{k}$$,$$\overline{b}=-\overline{i}+3\overline{j}+4\overline{k},\ \overline{c}=4\overline{i}-2\overline{j}-6\overline{k}$$ form the sides of the triangle then length of the median bisecting the vector $${c}$$ is

A
$$\sqrt{12}$$ units

B
$$\sqrt{6}$$ units

C
$$2\sqrt{6}$$ units

D
$$2\sqrt{3}$$ units

Solution

$$\vec a = 3\hat i+\hat j -2\hat k = \vec {BC}$$
$$\vec b = -\hat i+3\hat j+4\hat k=\vec{AC}$$
$$\vec c = 4\hat i-2\hat j-6\hat k = \vec{AB}$$

$$|\vec a|^2 = (3^2+1^2+2^2) = 14$$

$$|\vec b|^2 = 1^2+3^2+4^2 = 26$$

$$|\vec c|^2 = 4^2+2^2+6^2 = 56$$

Length of median through vertex $$C$$ passing through $$AC$$:
$$L = \dfrac12\sqrt{2(a^2+b^2)-c^2}$$

$$L = \dfrac12\sqrt{2(14+26) - 56} = \dfrac12\sqrt{24} = \sqrt6$$

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 23

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Vector Algebra Test - 23

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SHARING IS CARING

Question 1

$$\dfrac{\vec{a}}{|\vec{a}|}$$

$$\vec{a}|\vec{a}|$$

$$a^2$$

$$\hat{i}$$

Solution

Question 2

(9, -4)

(-9, -4)

(9, 4)

none of these

Solution

Question 3

(+1,+2)

(+1,-2)

(-1,+2)

(-1,-2)

Solution

Question 4

$$ -5\hat{i}-1\hat{j} $$

$$ \hat{i}-5\hat{j} $$

$$ \hat{i}+5\hat{j} $$

none of these

Solution

Question 5

$$\mid \vec a \mid^2 (\vec b \cdot \vec c)$$

$$\mid \vec b \mid^2 (\vec a \cdot \vec c)$$

$$\mid \vec c \mid^2 (\vec a \cdot \vec b)$$

none of these

Solution

Question 6

$$48 \hat b$$

$$- 48 \hat b$$

$$48 \hat a $$

$$48 \hat a$$

Solution

Question 7

$$\hat { BC } $$

$${ \dfrac { 1 }{ 2 } }\hat { BC } $$

$$ { \dfrac { 3 }{ 2 } \hat { BC } } $$

$$\dfrac { 3 }{ 4 } \hat { BC } $$

Solution

Question 8

$$3\hat{i}+2\hat{j}+\hat{k}$$

$$\displaystyle \frac{1}{2}(6\hat{i}+\hat{j}+\hat{k})$$

$$\displaystyle \frac{1}{2}(5\hat{i}+6\hat{j}+2\hat{k})$$

$${\dfrac{1}{2}(9\overline{i}}+7\overline{j}+3\overline{k})$$

Solution

Question 9

$$\vec{\mathrm{O}\mathrm{A}}$$

$$\vec{\mathrm{O}^{'}\mathrm{A}}$$

$$\vec{\mathrm{A}\mathrm{P}}$$

$$\vec{\mathrm{A}\mathrm{O}}$$

Solution

Question 10

$$\sqrt{12}$$ units

$$\sqrt{6}$$ units

$$2\sqrt{6}$$ units

$$2\sqrt{3}$$ units

Solution

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