Vector Algebra Test

Result

Vector Algebra Test - 75

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Question 1
1 / -0

If $$ \sum_{i=1}^{n} \vec{a}_{i}=\vec{0} $$ where $$ \left|\vec{a}_{i}\right|=1 \forall i, $$ then the value of $$ \sum_{1 \leq i<j \leq n} \vec{a}_{i} \cdot \vec{a}_{j} $$ is

A
$$-n/2$$

B
$$-n$$

C
$$n/2$$

D
$$n$$

Solution
Question 2
1 / -0

In parallelogram ABCD, $$|\overline { AB } |=a,|\overline { AD } |=b$$ and $$|\overline { AC } |=c$$ then $$\overline { DB } ,\overline { AB } $$ has the value

A
$$\frac {3a^2+b^2-c^2}{2}$$

B
$$\frac {3b^2+c^2-a^2}{2}$$

C
$$\frac {3c^2+b^2-a^2}{2}$$

D
$$\frac {a^2+b^2+c^2}{2}$$

Solution
Question 3
1 / -0

$$|\hat{i} \times p|^2 + |\hat{j} \times p|^2 +|\hat{k} \times p|^2$$ is equal to

A
$$4p^2$$

B
$$2p^2$$

C
$$p^2$$

D
$$\dfrac{p^2}{2}$$

Solution
Question 4
1 / -0

If the vectors $$\vec{a}$$ , $$\vec{b}$$ , $$\vec{c}$$ satisfying $$\vec{a}$$ + $$\vec{b}$$ + 2$$\vec{c}$$ = 0 . If $$\left | \vec{a} \right |$$ = 1 , $$\left | \vec{b} \right |$$ = 4 , $$\left | \vec{c} \right |$$ = 2 , then $$\vec{a}.\vec{b}$$ + $$\vec{b}.\vec{c}$$ + $$\vec{c}.\vec{a}$$ =

A
$$-\dfrac{7}{2}$$

B
$$-\dfrac{17}{2}$$

C
$$\dfrac{17}{2}$$

D
$$\dfrac{7}{2}$$

Solution
Question 5
1 / -0

A stone projected vertically upwards raises 's' feets in 't' seconds where $$_{ S }=112t-{ 16t }^{ 2 }$$ then the maximum height it reached is

A
195 ft

B
194 ft

C
196 ft

D
216 ft

Solution

$$s=112 t-16 t^{2}$$
for maximum height
$$\dfrac{d s}{d t}=0$$
$$\dfrac{d\left(112 t-16 t^{2}\right)}{d t}=0$$
$$\dfrac{d(112 t)}{d t}-\dfrac{d\left(16 t^{2}\right)}{d t}=0$$
$$112-16(2) t^{2-1}=0$$
$$112-32 t=0$$
$$32 t=112$$
$$t=\dfrac{112}{32}=\dfrac{28}{8}=\frac{7}{2}$$
$$\therefore$$ maximum height
$$=112\left(\dfrac{7}{2}\right)-16\left(\dfrac{7}{2}\right)^{2}$$
$$=56 \times 7-4 \times 49$$
$$=392-196$$
$$=196 \mathrm{ft}$$
Question 6
1 / -0

The position vector of A is $$2\vec { i } +3\vec { j } +4\vec { k } $$$$\vec { AB } =5\vec { i } +7\vec { j } +6\vec { k } $$, then the position vector of B is

A
$$-7\vec { i } -10\vec { j } -10\vec { k } $$

B
$$7\vec { i } -10\vec { j } +10\vec { k } $$

C
$$7\vec { i } +10\vec { j } -10\vec { k } $$

D
$$7\vec { i } +10\vec { j } +10\vec { k } $$
Question 7
1 / -0

The projection of the join of the point (3, 4, 2), (5, 1, 8) on the line whose d.c.'s are $$\left( \dfrac { 2 }{ 7 } ,-\dfrac { 3 }{ 7 } ,\dfrac { 6 }{ 7 } \right) $$ is

A
7

B
$$\left( \dfrac { 46 }{ 13 } \right) $$

C
$$\left( \dfrac { 42 }{ 13 } \right) $$

D
$$\left( \dfrac { 38 }{ 13 } \right) $$

Solution
Question 8
1 / -0

Let OAB be a regular triangle with side unity (o being otogin). Also M, N are the points of intersection of AB, M being closer to A and N closer to B. Position vectors of A, B, M and N are $$\vec { a } ,\vec { b } ,\vec { m } $$ and $$\vec { n } $$ respectively. Which of the following hold (s) good ?

A
$$\vec { m } =x\vec { a } +y\vec { b } \Rightarrow \dfrac { 2 }{ 3 } $$ and $$y=\dfrac { 1 }{ 3 } $$

B
$$\vec { m } =x\vec { a } +y\vec { b } \Rightarrow x=\dfrac { 5 }{ 6 } $$ and $$y=\dfrac { 1 }{ 6 } $$

C
$$\vec { m } .\vec { n } $$ equals $$\dfrac { 13 }{ 18 } $$

D
$$\vec { m } .\vec { n } $$ equals $$\dfrac { 15 }{ 18 } $$
Question 9
1 / -0

Given $$\overline { a } = x \hat { i } + y \hat { j } + 2 \hat { k } , \overline { b } = \hat { i } - \hat { j } + \hat { k } , \overline { c } = \hat { i } + 2 \hat { j } ;( \overline { a } \hat { } \overline { b } ) = \pi / 2 , \overline { a } .\overline { c } = 4$$ then

A
$$[ \overline { a } \overline { b } \overline { c } ] ^ { 2 } = \left| \overline { a } \right| $$

B
$$[ \overline { a } \overline { b } \overline { c } ]= \left| \overline { a } \right| $$

C
$$[ \overline { a } \overline { b } \overline { c } ]= 0$$

D
none of these

Solution
Question 10
1 / -0

If $$\bar { a } ,\bar { b } ,\bar { c } $$ are position vectors of the points A,B,C respectively such that $$9\bar { a } -7\bar { b } -2\bar { c } =\bar { 0 } $$ then point B divides AC in the ratio.....

A
Internally 7:2

B
Externally 9:2

C
Internally 9:7

D
Externally 2:7

Solution