Vector Algebra Test

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Vector Algebra Test - 60

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Question 1
1 / -0

Given a parallelogram $$ABCD$$. If $$|\vec{AB}=a, |\vec{AD}|=b$$ and $$|vec{AC}|=c$$, then $$|\vec{DB}|.|\vec{AB}|$$ has the

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

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

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

D
$$none$$

Solution
Question 2
1 / -0

Let $$\vec a$$ and $$\vec b$$ be two unit vectors such that $$\left| {\vec a + \vec b} \right| = \sqrt 3$$. If $$\vec c = \vec a + 2\vec b + 3(\vec a \times \vec b)$$, then $$2\left| {\vec c} \right|$$ is equal to:

A
$$\sqrt {55} $$

B
$$\sqrt {51} $$

C
$$\sqrt {43} $$

D
$$\sqrt {37} $$

Solution

$$ \left | \vec{a}+\vec{b} \right | = \sqrt{3}$$ [Given]
$$ \Rightarrow \left | \vec{a}+\vec{b} \right | = \left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+2\vec{a}\vec{b} = 3 $$
$$ \Rightarrow 2+2cos\theta = 3 $$
$$[ \therefore \vec{a}$$ $$ \vec{b}$$ are unit vector]
$$ cos\theta =1/2$$
$$ \Rightarrow \theta = \pi /3$$
$$ \Rightarrow $$ Now, $$ \vec{c} = \vec{a}+2\vec{b}+3(\vec{a}\times \vec{b})$$
$$ = \vec{a}+2\vec{b}+3\left | \vec{a} \right |\left | \vec{b} \right |sin\theta \,\hat{n}$$
$$=\vec{a}+2\vec{b}+\frac{3\sqrt{3}}{2}\hat{n}$$
$$\Rightarrow \left | \vec{c} \right |^{2}=\left | \vec{a} \right |^{2}+4\left | \vec{b} \right |^{2}+\frac{27}{4}\left | \vec{n} \right |^{2}+$$
$$4\vec{a}.\vec{b}+2\times 3\sqrt{3}\hat{n}.\vec{b}+3\sqrt{3}\hat{n}.\vec{a}$$
$$ [\because \vec{n}\perp \vec{b}$$ and $$ \vec{n}\perp \vec{a}]$$
$$ = 5+\frac{27}{4}+2 = \frac{28+27}{4} = \frac{55}{4}$$
$$ \Rightarrow \left | \vec{c} \right | = \frac{\sqrt{55}}{a}$$
$$ \Rightarrow 2\left | \vec{c} \right | = \sqrt{55}$$
Question 3
1 / -0

Let $$A(\overline { a } ),B(\overline { b } ), C(\overline { c } )$$ be the vertices of the triangle $$ABC$$ and let $$D,E,F$$ be the mid points of the sides $$BC,CA,AB$$ respectively. If $$P$$ divides the median $$AD$$ in the ratio $$2:1$$ then the position vector of $$P$$ is

A
$$\overline { 0 } $$

B
$$\overline { a } +\overline { b } +\overline { c } $$

C
$$\cfrac{\overline { b } +\overline { c } +\overline { d } }{3}$$

D
$$\cfrac{2\overline { a } +\overline { b } +\overline { c } }{3}$$

Solution

From section formula,
$$\vec{P}=\dfrac{2\left(\dfrac{\vec b+\vec c}{2}\right)+\vec a}{2H}=\dfrac{\vec{b}+\vec{c}+\vec{a}}{3}$$
$$\Rightarrow (C)$$
Question 4
1 / -0

If $$|\bar {a}-\bar {b}|=|\bar {a}|=|\bar {b}|$$, where $$\bar a$$ and $$\bar b$$ are non zero vecrors then the angle between $$\bar {a}-\bar {b}$$ and $$\bar b$$ is

A
$$120^{o}$$

B
$$45^{o}$$

C
$$60^{o}$$

D
$$90^{o}$$

Solution
Question 5
1 / -0

$$ABC$$ is an isosceles triangle right angled at $$A$$. Force of magnitude $$2\sqrt{2},5$$ and $$6$$ act along $$\overline { BC } ,\overline { CA } ,\overline { AB } $$ respectively. The magnitude of their resultant force is

A
$$4$$

B
$$5$$

C
$$11+2\sqrt{2}$$

D
$$30$$

Solution

REF.Image.
$$ AB = AC \Rightarrow \angle B = \angle C$$
And $$ \angle B+\angle C+90-180$$
$$ \Rightarrow \angle B = \angle C = 45^{\circ}$$
Taking components of $$ 2\sqrt{2}$$
$$ 2\sqrt{2}cos45 = 2\sqrt{2}\times \frac{1}{\sqrt{2}} = 2 $$
$$ 2\sqrt{2}sin45 = 2\sqrt{2}\times \frac{1}{\sqrt{2}} = 2 $$
$$ \Rightarrow $$ Resultant =$$ \sqrt{16+9} = 5 N $$
$$ \Rightarrow (B)$$
Question 6
1 / -0

The length of the projection of the line segment joining the points $$(5, -1, 4)$$ and $$(4, -1, 3)$$ on the plane , $$ x+ y+ z = 7$$ is :

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

B
$$\dfrac{1}{3}$$

C
$$\sqrt{\dfrac{2}{3}}$$

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

Solution

Here $$PQ$$ is the projection
Also $$PQ=BC$$
$$AC=\overline { AB } .\overline { BC } =\left( \hat { i } +\hat { k } \right) .\left( \dfrac { \hat { i } +\hat { j } +\hat { k } }{ \sqrt { 3 } } \right) $$
$$=2/\sqrt {3}$$
Also, $$AB=\sqrt {2}$$
$$AB=AC^{2}+BC^{2}=2=4/3+BC^{2}$$
$$BC^{2}=2/3$$
$$BC=\sqrt {2/3}$$
Question 7
1 / -0

$$\vec{a},\vec{b},\vec{c},\vec{d}$$ are the position vectors of four coplanar points A,B,C,D respectively. If no three of them are collinear and $$|\vec{a}-\vec{d}|=|\vec{b}-\vec{d}|=|\vec{c}-\vec{d}|$$ then for triangle ABC, D is

A
centroid

B
orthocenter

C
incenter

D
circumcenter
Question 8
1 / -0

If AD, BE and CF are $$\Delta ABC$$, then $$\\ \vec { AD } +\vec { BE } \vec { +CF } $$

A
$$\vec { 0 } $$

B
1

C
0

D
2

Solution
Question 9
1 / -0

Let $$\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}, b=\hat{i}+2\hat{j}-2\hat{k}$$. Then a unit vector perpendicular to both $$\vec{a}-\vec{b}$$ and $$\vec{a}+\vec{b}$$ is :

A
$$\dfrac{-1}{3}(-2\hat{i}+2\hat{j}+\hat{k})$$

B
$$\dfrac{1}{3}(-2\hat{i}+2\hat{j}-\hat{k})$$

C
$$\dfrac{1}{3}(2\hat{i}-2\hat{j}+\hat{k})$$

D
$$\dfrac{1}{3}(\hat{i}+\hat{j}+\hat{k})$$

Solution

We have,
$$ \vec { a } =3\hat { i } +2\hat { j } +2\hat { k } ,\, \, \vec { b } =\hat { i } +2\hat { j } -2\hat { k } \\ \vec { a } +\vec { b } =4\hat { i } +4\hat { j } \, \, \, \, \, \, \hat { a } -\hat { b } =2\hat { i } +4\hat { k } \\ \vec { d } =\left( { \vec { a } -\vec { b } } \right) \times \left( { \vec { a } \times \vec { b } } \right) $$

Therefore,
$$=\left| { \begin{array} { *{ 20 }{ c } }i & { \hat { j } } & { \hat { k } } \\ 2 & 0 & 4 \\ 4 & 4 & 0 \end{array} } \right| =-16\hat { i } +16\hat { j } +8\hat { k } $$

Unit vector perpendicular to
$$ \vec { a } -\vec { b } $$ and $$\vec { a } +\vec { b }$$ is
$$=\dfrac { { \left( { 16\hat { i } +16\hat { j } +8\hat { k } } \right) } }{ { \sqrt { 576 } } } =\dfrac { { -8\left( { -2\hat { i } +2\hat { j } +\hat { k } } \right) } }{ { 24 } } \\ =\dfrac { { -1 } }{ 3 } \left( { -2\hat { i } +2\hat { j } +\hat { k } } \right) $$

Hence, this is the answer.
Question 10
1 / -0

If $$\vec a,\vec b, \vec c$$ are three coplanar vectors, then $$v\left[ 2\vec { a } +3\vec { b } ,2\vec { b- } 5\vec { c } ,2\vec { c } +3\vec { a } \right]$$ is

A
$$0$$

B
$$1$$

C
$$-\sqrt 3$$

D
$$\sqrt 3$$

Solution