Vector Algebra Test

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Vector Algebra Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

The position vectors of two vertices and the centroid of a triangle are $$\vec{i}+\vec{j}$$, $$2\vec{i}-\vec{j}+\vec{k}$$ and $$\vec{k}$$ respectively, then the position vector of the third vertex of the triangle is

A
$$-3\vec{i}+2\vec{k}$$

B
$$3\vec{i}-2\vec{k}$$

C
$$\vec{i}+\dfrac{2}{3}\vec{k}$$

D
None of these

Solution

Let the third vertex be $$C=x\vec{i}+y\vec{j}+z\vec{k}$$
and $$A=(1,1,0)$$, $$B=(2,-1,1)$$, $$G=(0,0,1)$$
Hence $$G=\left (\dfrac{x_{1}+x_{2}+x_{3}}{3},\dfrac{y_{1}+y_{2}+y_{3}}{3},\dfrac{z_{1}+z_{2}+z_{3}}{3}\right)$$
$$(0,0,1)=\left (\dfrac{1+2+x}{3},\dfrac{1-1+y}{3},\dfrac{0+1+z}{3}\right)$$
$$(0,0,1)=\left (\dfrac{x+3}{3},\dfrac{y}{3},\dfrac{z+1}{3}\right)$$
Hence $$x+3=0$$ gives $$x=-3$$
$$\dfrac{y}{3}=0$$ gives $$y=0$$
And $$\dfrac{z+1}{3}=1$$ gives $$z=2$$
Hence $$C=-3\vec{i}+2\vec{k}$$.
Question 2
1 / -0

$$\displaystyle R(\bar{r})$$ is any point on a semi-circle, $$\displaystyle P(\bar{p})$$ and $$\displaystyle Q(\bar{q})$$ are the position vector of the end point of the diameter of that semi-circle, then $$\displaystyle \overline{PR}\cdot \overline{QR}$$ is equal to

A
$$1$$

B
$$0$$

C
$$3$$

D
Not defined

Solution

If any point on the semi-circle lies and we join that point to the end points of diameter then the angle formed between them is always normal or 90
So the angle between $$\vec{PR}\ and\ \vec{QR}$$ be 90
$$\vec{PR}\cdot\vec{QR}=\left |\vec{PR}\right | \left |\vec{QR}\right |\cos90$$
$$\vec{PR}\cdot\vec{QR}=\left |\vec{PR}\right | \left |\vec{QR}\right |\times0$$
$$\vec{PR}\cdot\vec{QR}=0$$
Question 3
1 / -0

In a triangle OAB, E is the mid-point of OB and D is a point on AB such that AD: DB = 2 : 1. If OD and AE intersect at P, determine the ratio OP : PD using vector methods.

A
$$OP:PD=2:3$$

B
$$OP:PD=3:2$$

C
$$OP:PD=1:3$$

D
$$OP:PD=3:1$$

Solution

With $$O$$ as origin let $$a$$ and $$b$$ be the position vectors of $$A$$ and $$B$$ respectively.
Then the position vector of $$E$$, the mid-point of $$OB$$, is $$\displaystyle\frac{b}{2}$$
Again, since $$AD: DB=2:1$$, the position vector of $$D$$ is $$\displaystyle \frac{1\cdot a+2b}{1+2}=\frac{a+2b}{3} $$
Equation of $$OD$$ and $$AE$$ are $$\displaystyle r=t\frac{a+2b}{3} $$ ...(1)
and $$\displaystyle r=a+s\left ( \frac{b}{2}-a \right )$$ or $$\displaystyle r=\left ( 1-s \right )a+s\frac{b}{2}$$ ...(2)
If they intersect at $$p$$, then we will have identical values of $$r$$.
Hence comparing the coefficients of $$a$$ and $$b$$, we get
$$\displaystyle \frac{t}{3}=1-s,\frac{2t}{3}=\frac{s}{2}\therefore t=\frac{3}{5}$$ or $$\displaystyle s=\frac{4}{5}.$$
Putting for t in (1) or for s in (2), we get the position vector of point of intersection $$P$$ as $$\displaystyle \frac{a+2b}{5}$$ ...(3)
Now let $$P$$ divide $$OD$$ in the ratio $$\displaystyle\lambda :1.$$
Hence by ratio formula the $$P.V.$$ of $$P$$ is $$\displaystyle \dfrac{\dfrac{\lambda \left ( a+2b \right )}{3}+1.0}{\lambda +1}=\dfrac{\lambda }{3\left(\lambda +1 \right )}\left ( a+2b \right )$$ ....(4)
Comparing (3) and (4), we get $$\displaystyle \frac{\lambda }{3\left ( \lambda +1 \right )}=\frac{1}{5}$$$$\Rightarrow\displaystyle 5\lambda =3\lambda +3\Rightarrow 2\lambda =3$$$$\displaystyle\Rightarrow \lambda =\frac{3}{2}$$
$$\therefore OP:PD=3:2$$
Question 4
1 / -0

In a triangle ABC, D divides BC in the ratio 3 : 2 and E divides CA in the ratio 1 : 3. The lines AD and BE meet at H and CH meets AB in F. Find the ratio in which F divides AB.

A
$$\displaystyle AF:FB=2:1$$

B
$$\displaystyle AF:FB=1:2$$

C
$$\displaystyle AF:FB=2:3$$

D
$$\displaystyle AF:FB=3:2$$

Solution

Take $$A$$ as origin and the position vectors of $$B$$ and $$C$$ be $$b$$ and $$c$$.
Hence the position vectors of other points under given conditions are
$$\displaystyle D=\dfrac{3c+2b}{5}, E=\dfrac{1.0+3c}{4}= \dfrac{3}{4}c.$$

Equations of $$AD$$ and $$BE$$ are
$$AD$$ is $$\displaystyle r= t\dfrac{3c+2b}{5}$$
$$BE$$ is $$\displaystyle r= (1- s )b+s\cdot\dfrac{3}{4}c.$$
They intersect at $$H$$.

Comparing coefficients of $$b$$ and $$c$$, we get
$$\displaystyle \dfrac{2}{5}t= 1-s, \dfrac{3}{5}t= \dfrac{3}{4}s.\\$$
$$\displaystyle \therefore s= \dfrac{4}{5}t.\\$$
$$\displaystyle\therefore \dfrac{2}{5}t+\dfrac{4}{5}t= 1\\$$
$$\displaystyle \therefore t= \dfrac{5}{6}, s=\dfrac{4}{6}\\$$
Point $$H$$ is $$\displaystyle \dfrac{3c+2b}{6}.$$

Now $$F$$ is point of inersection of $$AB$$ and $$CH$$ whose equations are $$r = t b$$
and $$\displaystyle r=\left ( 1-s \right )c+s\dfrac{3c+2b}{6}$$

Comparing the coefficients,
$$\displaystyle t=\dfrac{2s}{6}=\dfrac{s}{3}$$ and $$\displaystyle 1-s+\dfrac{3}{6}s=0\Rightarrow s=2\Rightarrow t=\dfrac{1}{3}s=\dfrac{2}{3}$$

$$\displaystyle \therefore$$ P.V. of $$\displaystyle F=\dfrac{2}{3}b$$ or $$\displaystyle \vec{AF}=\dfrac{2}{3}b,\vec{FB}=b-\dfrac{2}{3}b=\dfrac{1}{3}b$$

$$\therefore AF:FB=2:1$$
Question 5
1 / -0

The difference of the squares on the diagonals is four times the rectangle contained by either of these sides and the projection of the other upon it.

A
$$\displaystyle = 4$$ ( rectangle contained by AB and projection of AC on AB).

B
$$\displaystyle = 1$$ ( rectangle contained by AB and projection of AC on AB).

C
$$\displaystyle = 8$$ ( rectangle contained by AB and projection of AC on AB).

D
$$\displaystyle = 2$$ ( rectangle contained by AB and projection of AC on AB).

Solution

Again $$\displaystyle \vec{AD}^{2}-\vec{BC}^{2}= \left ( b+c \right )^{2}-\left ( c-b \right )^{2}= 4b.c$$
$$\displaystyle = 4\vec{AB}.\vec{AC}= 4AB$$ $$($$Projection of $$AC$$ on $$AB)$$
$$\displaystyle = 4$$ $$($$rectangle contained by $$AB$$ and projection of $$AC$$ on $$AB).$$
Question 6
1 / -0

$$ABC$$ is a $$\displaystyle \Delta $$ and $$G$$ is its centroid. If $$\displaystyle \overline{AB}= \bar{b}$$ and $$\displaystyle \overline{AC}= \bar{c}$$, then $$\displaystyle \overline{AG}$$ is equal to

A
$$\displaystyle \dfrac{2}{3}\left ( \bar{b}+\bar{c} \right )$$

B
$$\displaystyle \bar{b}+\bar{c} $$

C
$$\displaystyle \dfrac{1}{3}\left ( \bar{b}+\bar{c} \right )$$

D
$$\displaystyle \dfrac{3}{2}\left ( \bar{b}+\bar{c} \right )$$

Solution

Taking $$A$$ as origin.
So, position vector of $$\displaystyle B=\overline{AB}=\bar{b}$$
and position vector of $$\displaystyle C=\overline{AC}=\bar{c}$$
Hence by centroid formula,
$$\displaystyle \overline{AG}$$ = position vector of $$G$$ $$\displaystyle =\dfrac{\bar{a} + \bar{b}+\bar{c}}{3}$$
$$ = \dfrac{\bar{b}+\bar{c}}{3}$$
Question 7
1 / -0

$$\displaystyle \bar{a}= x\hat{i}+y\hat{j}+z\hat{k},\bar{b}= \hat{j}$$ then the vector $$\displaystyle \bar{c}$$ for which $$\displaystyle \bar{a},\bar{b},\bar{c}$$ form a right hand triad

A
$$\displaystyle x(\hat{i}-\hat{k})$$

B
$$\displaystyle \bar{0}$$

C
$$\displaystyle -z\hat{i}+x\hat{k}$$

D
$$\displaystyle y\hat{j}$$

Solution

Three vectors a, b and c are said to form a right handed system if a right threaded screw when rotated from a to b the screw moves in direction of c.
Even you can say that axb=c
$$\vec{c}= \begin{vmatrix}\hat{i}&\hat{j}&\hat{k} \\x&y&z\\0&1&0\end{vmatrix}$$
$$\vec{c}=\hat{i}(-z)-\hat{j}(0)+\hat{k}(x)$$
$$\vec{c}=-z\hat{i}+x\hat{k}$$
Question 8
1 / -0

Let $$\displaystyle \bar{p}$$ and $$\displaystyle \bar{q}$$ be two distinct points. Let $$R$$ and $$S$$ be the points dividing $$PQ$$ internally and externally in the ratio $$2:3$$. If $$\displaystyle \overline{OR} \perp \overline{OS},$$ then

A
$$\displaystyle 9p^{2}= 4q^{2}$$

B
$$\displaystyle 4p^{2}= 9q^{2}$$

C
$$\displaystyle 9p= 4q $$

D
$$\displaystyle 4p= 9q $$

Solution

Position vector of $$\displaystyle R=\dfrac{2\bar{q+3\bar{p}}}{5}$$
position vector of $$\displaystyle S=\dfrac{2\bar{q-3\bar{p}}}{\left ( -1 \right )}$$
$$\displaystyle =3\vec{p}-2\bar{q}$$
Since $$\displaystyle \overline{OR}\perp \overline{OS}$$,
$$\therefore \overline{OR}\cdot \overline{OS}=0$$
$$\displaystyle \therefore 9p^{2}=4q^{2}$$
Question 9
1 / -0

Let $$\bar{a}$$, $$\bar{b}$$ and $$\bar{c}$$ be vectors with magnitudes $$3, 4$$ and $$5$$ respectively and $$\bar{a}+\bar{b}+\bar{c}=0$$, then the value of $$\bar{a}.\bar{b}+\bar{b}.\bar{c}+\bar{c}.\bar{a}$$ is

A
$$48$$

B
$$-26$$

C
$$25$$

D
$$-25$$

Solution

We have
$$|a|= 3$$
$$|b|=4$$
$$|c|=5$$
Clearly (as $$c^{2}=a^{2}+b^{2}$$)
$$\bar{a}$$, $$\bar{b}$$ and $$\bar{c}$$ forms the right angle $$\triangle $$
$$\therefore $$ $$\bar{a}.\bar{b}=0$$
$$\therefore $$ $$\bar{a}.\bar{b}+\bar{b}.\bar{c}+\bar{c}.\bar{a}=\bar{c}.\left ( \bar{a}+\bar{b} \right )$$
$$=\bar{c}.\left ( -\bar{c} \right )$$
$$=-\left | \bar{c} \right |^{2}=-25$$
Question 10
1 / -0

The sides of a $$\triangle$$ are in A.P, then the line joining the centroid to the incenter is parallel to

A
The largest side

B
The middle side

C
The smallest side

D
None of these

Solution

$$G=(\dfrac{x_{1}+x_{2}+x_{3}}{3},\dfrac{y_{1}+y_{2}+y_{3}}{3})$$

$$=(\dfrac{bx_{1}+bx_{2}+bx_{3}}{3b},\dfrac{by_{1}+by_{2}+by_{3}}{3b})$$

$$I=(\dfrac{ax_{1}+bx_{2}+cx_{3}}{a+b+c},\dfrac{ay_{1}+by_{2}+cy_{3}}{a+b+c})$$

$$2b=(a+c)$$

Hence

$$I=(\dfrac{ax_{1}+bx_{2}+cx_{3}}{3b},\dfrac{ay_{1}+by_{2}+cy_{3}}{3b})$$

$$\vec{IG}=\dfrac{(b-a)x_{1}+(b-c)x_{2}}{3b}i+\dfrac{(b-a)y_{1}+(b-c)y_{2}}{3b}j$$

If
$$A=(x_{1},y_{1})$$
$$B=(x_{2},y_{2})$$
$$C=(x_{3},y_{3})$$
Then we can see that $$\vec{IG}$$ is parallel to the intermediate side.

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Vector Algebra Test - 34

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Vector Algebra Test - 34

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SHARING IS CARING

Question 1

$$-3\vec{i}+2\vec{k}$$

$$3\vec{i}-2\vec{k}$$

$$\vec{i}+\dfrac{2}{3}\vec{k}$$

None of these

Solution

Question 2

$$1$$

$$0$$

$$3$$

Not defined

Solution

Question 3

$$OP:PD=2:3$$

$$OP:PD=3:2$$

$$OP:PD=1:3$$

$$OP:PD=3:1$$

Solution

Question 4

$$\displaystyle AF:FB=2:1$$

$$\displaystyle AF:FB=1:2$$

$$\displaystyle AF:FB=2:3$$

$$\displaystyle AF:FB=3:2$$

Solution

Question 5

$$\displaystyle = 4$$ ( rectangle contained by AB and projection of AC on AB).

$$\displaystyle = 1$$ ( rectangle contained by AB and projection of AC on AB).

$$\displaystyle = 8$$ ( rectangle contained by AB and projection of AC on AB).

$$\displaystyle = 2$$ ( rectangle contained by AB and projection of AC on AB).

Solution

Question 6

$$\displaystyle \dfrac{2}{3}\left ( \bar{b}+\bar{c} \right )$$

$$\displaystyle \bar{b}+\bar{c} $$

$$\displaystyle \dfrac{1}{3}\left ( \bar{b}+\bar{c} \right )$$

$$\displaystyle \dfrac{3}{2}\left ( \bar{b}+\bar{c} \right )$$

Solution

Question 7

$$\displaystyle x(\hat{i}-\hat{k})$$

$$\displaystyle \bar{0}$$

$$\displaystyle -z\hat{i}+x\hat{k}$$

$$\displaystyle y\hat{j}$$

Solution

Question 8

$$\displaystyle 9p^{2}= 4q^{2}$$

$$\displaystyle 4p^{2}= 9q^{2}$$

$$\displaystyle 9p= 4q $$

$$\displaystyle 4p= 9q $$

Solution

Question 9

$$48$$

$$-26$$

$$25$$

$$-25$$

Solution

Question 10

The largest side

The middle side

The smallest side

None of these

Solution

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