Vector Algebra Test

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Vector Algebra Test - 33

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Question 1
1 / -0

$$P, Q$$ have position vectors $$\vec {a}$$ & $$\vec {b}$$ relative to the origin '$$O$$' and $$ X, Y$$ divide $$\vec{PQ}$$ intermally and extemally respectively in the ratio $$2:1$$, then vector $$\vec{XY}= $$

A
$$\displaystyle \:\dfrac{3}{2}(\vec b - \vec a)$$

B
$$\displaystyle \:\dfrac{4}{3}(\vec c - \vec a)$$

C
$$\displaystyle \:\dfrac{5}{6}\left ( \vec{b}- \vec{a}\right )$$

D
$$\displaystyle \:\dfrac{4}{3}(\vec b - \vec a)$$

Solution

$$X$$ divides $$\vec{PQ}$$ internally in $$2:1$$. Thus, $$\vec X = \dfrac{\vec a + 2\vec b}{2 + 1}$$
$$Y$$ divides $$\vec{PQ}$$ externally in $$2:1$$. Thus, $$\vec Y = \dfrac{2\vec b - \vec a}{2 - 1}$$
$$\vec X = \dfrac{\vec a + 2\vec b}{3}$$ and $$\vec Y = 2\vec b - \vec a$$
$$\therefore \vec{XY} = \vec Y - \vec X = \dfrac{6\vec b - 3\vec a -\vec a -2\vec b}{3}$$
$$\vec{XY}= \dfrac{4\vec b - 4\vec a}{3}$$
Question 2
1 / -0

If . and $$\times $$ represent dot product and cross product respectively then which of the following is meaningless?

A
$$\left ( \vec{a}\times \vec{b} \right ).\left ( \vec{c}\times \vec{d} \right )$$

B
$$\left ( \vec{a}\times \vec{b} \right )\times \left ( \vec{c}\times \vec{d} \right )$$

C
$$\left ( \vec{a}.\vec{b} \right ) .\left ( \vec{c}\times \vec{d} \right )$$

D
$$\left ( \vec{a}.\vec{b} \right )\times \left ( \vec{c}\times \vec{d} \right )$$

Solution

Cross product can only occur for two vectors.
$$a.b$$ is just a scalar quantity.
Hence the option $$D$$ is meaningless.
Question 3
1 / -0

If $$\vec{a}.\vec{b}=0$$ and $$\vec{a}\times \vec{b}=0$$ then

A
$$\vec{a}\parallel \vec{b}$$

B
$$\vec{a}\perp \vec{b}$$

C
$$\vec{a}=\vec{0}$$ or $$\vec{b}=\vec{0}$$

D
None of these

Solution

$$\vec{a}.\vec{b} = 0$$ and $$\vec{a} \times \vec{b} = 0$$

$$\vec{a}.\vec{b} = ab \cos{\theta}$$ and $$|\vec{a} \times \vec{b}| = ab \sin{\theta}$$

Since both of them are simultaneously $$0$$ and $$\cos{\theta}, \sin{\theta}$$ cannot be simultaneously $$0$$, we get either $$a$$ or $$b$$ has to be $$0$$.
Question 4
1 / -0

A straight line $$r=a+\lambda b$$ meets the plane $$r.n=0$$ in $$P$$. The position vector of $$P$$ is

A
$$\displaystyle a+\frac { a.n }{ b.n } b$$

B
$$\displaystyle a+\frac { b.n }{ a.n } b$$

C
$$\displaystyle a-\frac { a.n }{ b.n } b$$

D
None of these

Solution

A straight line $$r=a+\lambda b$$ meets the plane $$r.n=0$$ in $$P$$ for which $$\lambda$$ is given by
$$\displaystyle \left( a+\lambda b \right) .n=0\Rightarrow \lambda =-\frac { a.n }{ b.n } $$
Thus, the position vector of $$P$$ is
$$\displaystyle r=a-\frac { a.n }{ b.n } b$$ $$[$$ putting the value of $$\lambda$$ in $$r=a+\lambda b]$$
Question 5
1 / -0

$$P$$ is a point on the line through the point $$A$$ whose position vector is $$\overrightarrow{a}$$ and the line is parallel to the vector $$\overrightarrow{b}$$. If $$PA=6$$, the position vector of $$P$$ is

A
$$\overrightarrow{a}+6\overrightarrow{b}$$

B
$$\displaystyle \overrightarrow{a}+\dfrac{6}{\left |\overrightarrow{b} \right |}\overrightarrow{b}$$

C
$$\overrightarrow{a}-6\overrightarrow{b}$$

D
$$\displaystyle \overrightarrow{b}+\dfrac{6}{\left |\overrightarrow{a} \right |}\overrightarrow{a}$$

Solution

Line is parallel to vector $$\overrightarrow{b}$$
The required vector is at a distance of 6 units from $$\overrightarrow{a}$$
So, the required position vector will be $$\overrightarrow{a} + \dfrac{6}{|\overrightarrow{b}|} \overrightarrow{b}$$ since we need a unit vector in the direction of $$\overrightarrow{b}$$
Question 6
1 / -0

If $$G$$ and $$G'$$ be the centroids of the triangles $$ABC$$ and $$A'B'C'$$ respectively, then $$\overrightarrow { AA' } +\overrightarrow { BB' } +\overrightarrow { CC' } =$$

A
$$\displaystyle \dfrac { 2 }{ 3 } \overrightarrow { GG' } $$

B
$$\overrightarrow { GG' } $$

C
$$2\overrightarrow { GG' } $$

D
$$3\overrightarrow { GG' } $$

Solution

Let P.V. of $$A\left( \overrightarrow { a } \right) ,B\left( \overrightarrow { b } \right) ,C\left( \overrightarrow { c } \right) ,A'\left( \overrightarrow { { a }_{ 1 } } \right) ,B'\left( \overrightarrow { { b }_{ 1 } } \right) ,C'\left( \overrightarrow { { c }_{ 1 } } \right) $$

Now P.V. of $$\displaystyle G=\dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 3 } $$;

Similarly $$\displaystyle G'=\dfrac { \overrightarrow { { a }_{ 1 } } +\overrightarrow { { b }_{ 1 } } +\overrightarrow { { c }_{ 1 } } }{ 3 } $$

Now $$\overrightarrow { AA' } =\overrightarrow { { a }_{ 1 } } -\overrightarrow { a } ,\overrightarrow { BB' } =\overrightarrow { { b }_{ 1 } } -\overrightarrow { b } ,\overrightarrow { CC' } =\overrightarrow { { c }_{ 1 } } -\overrightarrow { c } $$

So $$\overrightarrow { AA' } +\overrightarrow { BB' } +\overrightarrow { CC' } =\left( \overrightarrow { { a }_{ 1 } } +\overrightarrow { { b }_{ 1 } } +\overrightarrow { { c }_{ 1 } } \right) -\left( \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } \right) $$

$$\displaystyle =3\left( \dfrac { \overrightarrow { { a }_{ 1 } } +\overrightarrow { { b }_{ 1 } } +\overrightarrow { { c }_{ 1 } } }{ 3 } -\dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 3 } \right) =3\overrightarrow { GG' } $$
Question 7
1 / -0

Five coplanar forces of equal magnitudes $$10 N$$ each, act at a point such that the angle between any two consecutive forces is same. The magnitude of their resultant is :

A
$$0$$

B
$$10 N$$

C
$$20 N$$

D
$$10 \sqrt {2} N$$

Solution

As the angle between any two consecutive forces have the same value $${ =72}^{\circ}$$ and all the forces are of equal magnitude. Hence their resultant is zero.
Question 8
1 / -0

Let $$\overrightarrow{AB}=3\hat{i}+\hat{j}-\hat{k}$$ and $$\overrightarrow{AC}=\hat{i}-\hat{j}+3\hat{k}$$. If the point $$P$$ on the line segment $$BC$$ is equidistant from $$AB$$ and $$AC$$ then $$\overrightarrow{AP}$$ is

A
$$2\hat{i}-\hat{k}$$

B
$$\hat{i}-2\hat{k}$$

C
$$2\hat{i}+\hat{k}$$

D
None of these

Solution

Since $$P$$ is equidistant from $$AC\ \&\ BC,$$ it lies on the angle bisector of $$\angle{BAC}$$
Let us also assume that $$A$$ is the origin.

So, $$\overrightarrow{P} = x\dfrac{3\hat{i} + \hat{j} - \hat{k} + \hat{i} - \hat{j} + 3\hat{k}}{\sqrt{11}}$$

$$\Rightarrow \overrightarrow{P} = y{4\hat{i} + 2\hat{k}}$$

Hence we can see the options to verify.
Question 9
1 / -0

If the vectors $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ and $$\vec{d}$$ are coplanar, then $$\left ( \vec{a}\times \vec{b} \right )\times \left ( \vec{c}\times \vec{d} \right )$$ is equal to

A
$$\vec{a}+\vec{b}+\vec{c}+\vec{d}$$

B
$$\vec{0}$$

C
$$\vec{a}+\vec{b}=\vec{c}+\vec{d}$$

D
None of these

Solution

Since $$\vec{a}, \vec{b}, \vec{c}, \vec{d}$$ are coplanar, the cross products $$\vec{a} \times \vec{b}$$ and $$\vec{c} \times \vec{d}$$ is a vector same for both.
$$\Rightarrow$$ their cross product will become 0 since the sine of angle between becomes 0.
Question 10
1 / -0

There are $$N$$ co-planar vectors each of magnitude $$V$$. Each vector is inclined to the preceding vector at angle $$2 \pi/N$$. What is the magnitude of their resultant?

A
zero

B
$$V/N$$

C
$$V$$

D
$$NV$$

Solution

When vectors are arranges at an angle $$\dfrac { 2\pi }{ N } $$ to the preceding vectors, they form an N-sided closed polygon. Adding all the sides of a polygon we get the resultant to be 0.

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Vector Algebra Test - 33

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Vector Algebra Test - 33

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Question 1

$$\displaystyle \:\dfrac{3}{2}(\vec b - \vec a)$$

$$\displaystyle \:\dfrac{4}{3}(\vec c - \vec a)$$

$$\displaystyle \:\dfrac{5}{6}\left ( \vec{b}- \vec{a}\right )$$

$$\displaystyle \:\dfrac{4}{3}(\vec b - \vec a)$$

Solution

Question 2

$$\left ( \vec{a}\times \vec{b} \right ).\left ( \vec{c}\times \vec{d} \right )$$

$$\left ( \vec{a}\times \vec{b} \right )\times \left ( \vec{c}\times \vec{d} \right )$$

$$\left ( \vec{a}.\vec{b} \right ) .\left ( \vec{c}\times \vec{d} \right )$$

$$\left ( \vec{a}.\vec{b} \right )\times \left ( \vec{c}\times \vec{d} \right )$$

Solution

Question 3

$$\vec{a}\parallel \vec{b}$$

$$\vec{a}\perp \vec{b}$$

$$\vec{a}=\vec{0}$$ or $$\vec{b}=\vec{0}$$

None of these

Solution

Question 4

$$\displaystyle a+\frac { a.n }{ b.n } b$$

$$\displaystyle a+\frac { b.n }{ a.n } b$$

$$\displaystyle a-\frac { a.n }{ b.n } b$$

None of these

Solution

Question 5

$$\overrightarrow{a}+6\overrightarrow{b}$$

$$\displaystyle \overrightarrow{a}+\dfrac{6}{\left |\overrightarrow{b} \right |}\overrightarrow{b}$$

$$\overrightarrow{a}-6\overrightarrow{b}$$

$$\displaystyle \overrightarrow{b}+\dfrac{6}{\left |\overrightarrow{a} \right |}\overrightarrow{a}$$

Solution

Question 6

$$\displaystyle \dfrac { 2 }{ 3 } \overrightarrow { GG' } $$

$$\overrightarrow { GG' } $$

$$2\overrightarrow { GG' } $$

$$3\overrightarrow { GG' } $$

Solution

Question 7

$$0$$

$$10 N$$

$$20 N$$

$$10 \sqrt {2} N$$

Solution

Question 8

$$2\hat{i}-\hat{k}$$

$$\hat{i}-2\hat{k}$$

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

None of these

Solution

Question 9

$$\vec{a}+\vec{b}+\vec{c}+\vec{d}$$

$$\vec{0}$$

$$\vec{a}+\vec{b}=\vec{c}+\vec{d}$$

None of these

Solution

Question 10

zero

$$V/N$$

$$V$$

$$NV$$

Solution

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