Vector Algebra Test

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Vector Algebra Test - 12

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Question 1
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If $$|\overrightarrow{C}|^2=60$$ and $$\overrightarrow{C} \times (\widehat{i}+2\widehat{j}+5\widehat{k})=\overrightarrow{0}$$, then a value of $$\overrightarrow{C}\cdot (-7 \widehat{i}+2\widehat{j}+3\widehat{k})$$ is :

A
$$4\sqrt{2}$$

B
$$24$$

C
$$12\sqrt{2}$$

D
$$12$$

Solution

$$ \vec {C}$$ is parallel to $$\widehat{i}+2\widehat{j}+5\widehat{k}$$.
$$ \Rightarrow \vec {C}=t(\widehat{i}+2\widehat{j}+5\widehat{k})$$, where '$$t$$' is a scalar.
$$ |\vec{C}|^2=60$$
$$\Rightarrow t^2 ( \hat{i}.\hat{i} + 4 \hat{j}.\hat{j} + 25 \hat{k}.\hat{k} ) = 60 $$
$$ \Rightarrow 60 = t^2 \times 30 $$
$$\Rightarrow t= \pm \sqrt{2}$$
Now
$$ \vec{C}.(-7i+2\widehat{j}+3\widehat{k})= \pm \sqrt{2} (i + 2 \widehat{j}+5\widehat{k})\cdot (-7\widehat{i} +2\widehat{j}+3 \widehat{k} )$$
$$=\pm \sqrt{2}[-7+4+15]$$
$$= \pm 12 \sqrt{2}$$
Question 2
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Let $$ABC$$ be a triangle whose circumcentre is at P. If the position vectors of $$A, B, C$$ and P are $$\vec {a}, \vec {b}, \vec {c}$$ and $$\dfrac {\vec {a} + \vec {b} + \vec {c}}{4}$$ respectively, then the position vector of the orthocentre of this triangle, is:

A
$$-\left (\dfrac {\vec {a} + \vec {b} + \vec {c}}{2}\right )$$

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

C
$$\dfrac {(\vec {a} + \vec {b} + \vec {c})}{2}$$

D
$$\vec {0}$$

Solution

O is orthocentre and G is centroid and C is circumcentre. G divides OC in $$2:1$$ ratio.
Position vector of centriod $$\vec {G} = \dfrac {\vec {a} + \vec {b} + \vec {c}}{3}$$
Position vector of circum center $$\vec {C} = \dfrac {\vec {a} + \vec {b} + \vec {c}}{4}$$
Apply Section Formula,
$$\vec {G} = \dfrac {2\vec {C} + \vec {R}}{3}$$
$$3\vec {G} = 2\vec {C} + \vec {R}$$
$$\vec {R} = 3\vec {G} - 2\vec {C} = (\vec {a} + \vec {b} + \vec {c}) - 2 \left (\dfrac {\vec {a} + \vec {b} + \vec {c}}{4}\right )$$
$$= \dfrac {\vec {a} + \vec {b} + \vec {c}}{2}$$
Question 3
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The vectors $$\overrightarrow{AB} = 3\hat{i} + 4\hat{k}$$ and $$\overrightarrow{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$$ are the sides of a triangle $$ABC$$, then the length of the median through $$A$$ is:

A
$$\sqrt{72}$$

B
$$\sqrt{33}$$

C
$$\sqrt{45}$$

D
$$\sqrt{18}$$

Solution

Take A as the origin(0,0)

so $$\vec {AB}= 3\hat i + 4\hat k$$ and $$AC = 5\hat i – 2\hat j + 4\hat k$$ will become position vector.

So coordinate of B is $$(3,0,4)$$ and C is$$(5,-2,4)$$.

The mid point of BC can be easily found by midpoint formula of two points as $$D(4,-1,4).$$

so length of median through $$A$$ is $$AD$$$$=\sqrt{(4-0)^2+(-1-0)^2+(4-0)^2}$$
$$=\sqrt{33}$$
Question 4
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If C is the mid point of AB and P is any point outside AB, then

A
$$\vec {\mathrm{P}\mathrm{A}}+\vec {\mathrm{P}\mathrm{B}}=2\vec {\mathrm{P}\mathrm{C}}$$

B
$$\vec {\mathrm{P}\mathrm{A}}+\vec {\mathrm{P}\mathrm{B}}=\vec {\mathrm{P}\mathrm{C}}$$

C
$$\vec {\mathrm{P}\mathrm{A}}+\vec {\mathrm{P}\mathrm{B}}+2\vec {\mathrm{P}\mathrm{C}}=0$$

D
$$\vec {\mathrm{P}\mathrm{A}}+\vec {\mathrm{P}\mathrm{B}}=\vec {\mathrm{P}\mathrm{C}}=0 $$

Solution

$$\overrightarrow { PA } +\overrightarrow { AC } +\overrightarrow { CP } =0$$
$$\overrightarrow { PB } +\overrightarrow { BC } +\overrightarrow { CP } =0$$
Adding we get
$$\overrightarrow { PA } +\overrightarrow { PB } +\overrightarrow { AC } +\overrightarrow { BC } +2\overrightarrow { CP } =0$$
Since $$\overrightarrow { AC } =-\overrightarrow { BC }$$ and $$\overrightarrow { CP } =-\overrightarrow { PC } $$
$$\Rightarrow \overrightarrow { PA } +\overrightarrow { PB } -2\overrightarrow { PC } =0$$
Question 5
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In a parallelogram ABCD, $$|\overrightarrow{AB}| = a, |\overrightarrow{AD}| = b$$ and $$|\overrightarrow{AC}| = c$$, then $$\overrightarrow{DB}.\overrightarrow{AB}$$ has the value

A
$$\displaystyle \frac{1}{2} (a^2 + b^2 + c^2)$$

B
$$\displaystyle \frac{1}{4} (a^2 + b^2 - c^2)$$

C
$$\displaystyle \frac{1}{3} (b^2 + c^2 - a^2)$$

D
$$\displaystyle \frac{1}{2} (a^2 + b^2 - c^2)$$

Solution

In a parallelogram ABCD, $$|\overrightarrow{AB}| = a, |\overrightarrow{AD}| = b$$ and $$|\overrightarrow{AC}| = c$$

$$c^2=a^2+b^2-2ab\cos B$$

$$\Rightarrow 2ab\cos B =-c^2+a^2+b^2$$

$$\vec{DB}\cdot \vec {AB}=(\vec {DA} + \vec {DC})\cdot\vec {AB}= ba \cos B + a^2=\displaystyle\frac{1}{2}(3a^2+b^2-c^2)$$

Hence, option D.
Question 6
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Let $$P,\ Q,\ R$$ and $$S$$ be the points on the plane with position vectors $$-2\hat{i}-\hat{j},\ 4\hat{i},\ 3\hat{i}+3\hat{j}$$ and $$-3\hat{i}+2\hat{j}$$ respectively. The quadrilateral $$PQRS$$ must be a

A
parallelogram, which is neither a rhombus nor a rectangle

B
square

C
rectangle, but not a square

D
rhombus, but not a square

Solution

Evaluating midpoint of $$PR$$ and $$QS$$ which
gives $$M\equiv[\dfrac{\hat{i}}{2}+\hat{j}]$$, same for both.
$$
\vec{PQ}=\vec{SR}=6\hat{i}+\hat{j}
$$
$$
\vec{PS}=\vec{QR}=-\hat{i}+3\hat{j}
$$

$$
\Rightarrow \vec{PQ}\cdot\vec{PS}\neq 0
$$
$$\vec{PQ}\Vert\vec{SR},\ \vec{PS}\Vert\vec{QR}$$ and $$|\vec{PQ}|=|\vec{SR}|,\ |\vec{PS}|=|\vec{QR}|$$
Hence, $$PQRS$$ is a parallelogram but not rhombus or rectangle.
Question 7
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Let $$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}$$ and $$\vec{c}=\hat{i}-\hat{j}-\hat{k}$$ be three vectors. A vector $$\vec{v}$$ in the plane of $$\vec{a}$$ and $$\vec{b}$$ , whose projection on $$\vec{c}$$ is $$\displaystyle \dfrac{1}{\sqrt{3}}$$ , is given by $$;$$

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

B
$$-3\hat{i}-3\hat{j}-\hat{k}$$

C
$$3\hat{i}-\hat{j}+3\hat{k}$$

D
$$\hat{i}+3\hat{j}-3\hat{k}$$

Solution

Let $$\overrightarrow { v } =\lambda \overrightarrow { a } +\mu \overrightarrow { b } $$
$$\Rightarrow\overrightarrow { v } =(\lambda +\mu )\hat { i } +(\lambda -\mu )\hat { j } +(\lambda +\mu )\hat { k } $$

Projection of $$\overrightarrow { v }$$ on $$\displaystyle\overrightarrow { c }=\dfrac { \overrightarrow { v } .\overrightarrow { c } }{ \left| \overrightarrow { c } \right| } = \dfrac { 1 }{ \sqrt { 3 } } $$

$$\Rightarrow\displaystyle\dfrac { (\lambda +\mu )-(\lambda -\mu )-(\lambda +\mu ) }{ \sqrt { 3 } } =\dfrac { 1 }{ \sqrt { 3 } }$$
$$\Rightarrow\mu -\lambda =1$$
or $$\mu =\lambda +1$$
$$\Rightarrow\overrightarrow { v } =(2\lambda +1)\hat { i } -\hat { j } +(2\lambda +1)\hat { k } $$
For $$ \lambda =1, \overrightarrow { v }=3\hat { i } -\hat { j }+3\hat { k }$$
Question 8
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The triangle $$ABC$$ is defined by the vertices $$A= (0,7,10)$$ , $$B=(-1,6,6)$$ and $$C=(-4,9,6)$$. Let $$D$$ be the foot of the attitude from $$B$$ to the side $$AC$$ then $$BD$$ is

A
$$\overline{i}+2\overline{j}+2\overline{k}$$

B
$$-\overline{i}+2\overline{j}+2\overline{k}$$

C
$$\overline{i}+2\overline{j}-2\overline{k}$$

D
$$\overline{i}-2\overline{j}+2\overline{k}$$

Solution

Given : $$A=(0,7,10), B=(-1,6,6)$$ and $$C=(-4,9,6)$$
Position vector(P.V.) of $$AB$$ is $$(-1,-1,-4)$$
$$\implies |AB|=\sqrt{(-1)^2+(-1)^2+(-4)^2}=\sqrt{18}$$
P.V. of $$BC$$ is $$(-3,3,0)$$
$$\implies |BC|=\sqrt{(-3)^2+3^2+0}=\sqrt{18}$$
P.V. of $$AC$$ is $$(-4,2,-4)$$
$$\implies |AC|=\sqrt{(-4)^2+2^2+(-4)^2}=6$$
Now, $$D$$ is the midpoint of $$AC$$
$$\implies D=\left(\dfrac{0-4}{2}, \dfrac{7+9}{2}, \dfrac{10+6}{2}\right)= (-2,8,8)$$
$$\implies \overrightarrow{BD} = -\hat{i}+2\hat{j}+2\hat{k}$$
Question 9
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The point $$C=(\dfrac{12}{5}, \dfrac{-1}{5},\dfrac{4}{5})$$ divides the line segment $$AB$$ in the ratio $$3:2$$. If $$B=(2,-1,2)$$ then $$A$$ is

A
$$

(3, 1, 1)$$

B
$$(3, 1,-1)$$

C
$$(3,-1,-1)$$

D
$$(-3,1,-1)$$

Solution

By Section Formula
$$C= \dfrac{3B+2A}{5}$$
$$(12,-1,4)= (6,-3,6)+2(A)$$
$$(6,2,-2)=2A$$
$$A=(3,1,-1)$$
Question 10
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$$ABCD$$ is a parallelogram and $$AC, BD$$ be its diagonals Then $$ \vec{AC} +\vec{BD}$$ is

A
$$2\vec{AB} $$

B
$$2\vec{BC} $$

C
$$3\vec{AB} $$

D
$$3\vec{BC} $$

Solution

$$ABCD$$ is a parallelogram so, $$AB||CD$$ and $$AD||BC$$
$$\vec{AC}=\vec{AB}+\vec{BC}$$ (addition Theorem)
$$\vec{BD}=\vec{BA}+\vec{AD}$$ (addition Theorem)
$$\vec{AC}+\vec{BD}=\vec{AB}+\vec{BC}+\vec{BA}+\vec{AD}$$
$$=\vec{AB}+\vec{BC}-\vec{AB}+\vec{BC}$$
$$=2\vec{BC}$$