Vector Algebra Test

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Vector Algebra Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

$$\vec{a} = 2 \widehat{i} - \widehat{j} + \widehat{k}, \vec{b} = \widehat{i} + 2\widehat{j} - \widehat{k} $$ and $$ \vec{c} = \widehat{i} + \widehat{j} - 2 \widehat{k}$$. A vector coplanar with $$\vec{b}$$ and $$\vec{c}$$ whose projection on $$\vec{a}$$ is magnitude $$\displaystyle \sqrt{\dfrac{2}{3}} $$ is

A
$$2 \widehat{i} + 3 \widehat{j} - 3 \widehat{k}$$

B
$$- 2 \widehat{i} - \widehat{j} + 5 \widehat{k}$$

C
$$2 \widehat{i} + 3 \widehat{j} + 3 \widehat{k}$$

D
$$2 \widehat{i} + \widehat{j} + 5 \widehat{k}$$

Solution

Let the required vector be $$\vec{r}$$, then
$$\vec{r} = x_1 \vec{b} + x_2 \vec{c}$$ and $$\vec{r} \cdot \vec{a} = \displaystyle \sqrt{\dfrac{2}{3}} (|\vec{a}|) = 2$$

Now, $$\vec{r} \cdot \vec{a} = x_1 \vec{a} \cdot \vec{b} + x_2 \vec{a} \cdot \vec{c}$$
$$\Rightarrow 2 = x_1 (2 - 2 - 1) + x_2 (2 - 1 - 2)$$
$$\Rightarrow x_1 + x_2 = -2$$
$$\Rightarrow \vec{r} = x_1 (\widehat{i} + 2 \widehat{j} - \widehat{k}) + x_2 (\widehat{i} + \widehat{j} - 2 \widehat{k})$$
$$= \widehat{i} (x_1 + x_2) + \widehat{j} (2x_1 + x_2) - \widehat{k} (2x_2 + x_1)$$
$$= - 2\widehat{i} + \widehat{j} (x_1 - 2) - \widehat{k} (-4 - x_1)$$,
where $$x_1 \in R$$
Question 2
1 / -0

'$$P$$' is a point inside the triangle $$ABC$$, such that $$\displaystyle BC\left ( \vec{PA} \right )+CA\left ( \vec{PB} \right )+AB\left ( \vec{PC} \right )=0,$$ then for the triangle $$ABC$$ the point $$P$$ is its :

A
Incentre

B
Circumcentre

C
Centroid

D
Orthocentre

Solution

Given that, $$P$$ is a point inside the triangle $$ABC$$, such that $$\displaystyle BC\left ( \vec{PA} \right )+CA\left ( \vec{PB} \right )+AB\left ( \vec{PC} \right )=0$$

Let, $$BC=a,CA=b,AB=c$$ and $$\vec{OA}=\bar{a},\vec{OB}=\bar{b},\vec{OC}=\bar{c}$$

$$\Rightarrow a(\bar{a}-\vec{OP})+b(\bar{b}-\vec{OP})+c(\bar{c}-\vec{OP})=0$$

$$\Rightarrow \vec{OP}=\displaystyle\dfrac{a\bar{a}+b\bar{b}+c\bar{c}}{a+b+c}$$

$$\therefore P$$ is an Incentre of triangle $$ABC$$.

Hence, option $$A$$.
Question 3
1 / -0

Let the pairs $$\vec{a}, \vec{b}$$ and $$\vec{c}, \vec{d}$$ each determine a plane, then the planes are parallel if

A
$$(\vec{a} \times \vec{c}) \times (\vec{b} \times \vec{d}) = \vec{0}$$

B
$$(\vec{a} \times \vec{c}) \cdot (\vec{b} \times \vec{d}) = \vec{0}$$

C
$$(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0}$$

D
$$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = \vec{0}$$

Solution

$$\vec{a} \times \vec{b}$$ is a vector perpendicular to the plane contining $$\vec{a}$$ and $$\vec{b}$$.

Similarly, $$\vec{c} \times \vec{d}$$ is a vector perpendicular to the plane containing $$\vec{c}$$ and $$\vec{d}$$.

Thus, the two planes will be parallel if their normals,
i.e., $$\vec{a} \times \vec{b}$$ and $$\vec{c} \times \vec{d}$$, are parallel.

Thus,
$$(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0}$$

Hence, option $$C$$.
Question 4
1 / -0

Let $$\vec{a},\vec{b},\vec{c}$$ be vectors of length $$3,4,5$$ respectively. Let $$\vec{a}$$ be perpendicular to $$\vec{b}+\vec{c},\vec{b}\,to\,\vec{c}+\vec{a}$$ and $$\vec{c}\,to\,\vec{a}+\vec{b}$$. Then $$\begin{vmatrix}\vec{a}+\vec{b}+\vec{c}\end{vmatrix}$$ is

A
$$\;2\sqrt{5}$$

B
$$\;2\sqrt{2}$$

C
$$\;10\sqrt{5}$$

D
$$\;5\sqrt{2}$$

Solution

$$\begin{vmatrix}\vec{a}+\vec{b}+\vec{c}\end{vmatrix}=\sqrt{(\vec{a}+\vec{b}+\vec{c})^2} =\sqrt{a^2+b^2+c^2+2\begin{pmatrix}\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}\end{pmatrix}} $$ {Modulus formula for vectors}
$$\because\,\vec{a}.\begin{pmatrix}\vec{b}+\vec{c}\end{pmatrix}=0,\;\vec{b}.\begin{pmatrix}\vec{c}+\vec{a}\end{pmatrix}=0\,\&\,\vec{c}.\begin{pmatrix}\vec{a}+\vec{b}\end{pmatrix}=0$$ [As the given conditions of being perpendicular}
$$\Rightarrow \vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}=0$$ {expanding the previous expression and substituting in the first expression}
$$\Rightarrow \begin{vmatrix}\vec{a}+\vec{b}+\vec{c}\end{vmatrix}=\sqrt{a^2+b^2+c^2}=\sqrt{3^2+4^2+5^2}=5\sqrt{2}$$
Question 5
1 / -0

$$ABCDEF$$ is a regular hexagon . The centre of hexagon is a point O. Then the value of
$$\overrightarrow{AB}+ \overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}$$ is

A
$$2\overrightarrow{AO}$$

B
$$4\overrightarrow{AO}$$

C
$$6\overrightarrow{AO}$$

D
Zero

Solution

We have
$$\overrightarrow{AB}+ \overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}$$
We have
$$\overrightarrow{AB}+ \overrightarrow{AC}=\overrightarrow{AB}+ (\overrightarrow{AB}+\overrightarrow{BC})=2\overrightarrow{AB}+\overrightarrow{BC}.............(i)$$
and
$$\overrightarrow{AE}+\overrightarrow{AF}=(\overrightarrow{AD}+\overrightarrow{DE})+(\overrightarrow{AD}+\overrightarrow{DE}+\overrightarrow{EF})=2\overrightarrow{AD}+2\overrightarrow{DE}+\overrightarrow{DF}............(ii)$$
From $$i$$ and $$ii$$ we see that $$\overrightarrow{AB}$$ and $$\overrightarrow{AC}$$ are vectorially opposite to $$\overrightarrow{DE}$$ and $$\overrightarrow{DF}$$
Thus we get
$$\overrightarrow{AB}+ \overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}= 3\overrightarrow{AD}=6\overrightarrow{AO}$$
Question 6
1 / -0

The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors a, b, c such that a.b = b.c = c.a = 1/2. What is the volume of the parallelopipe.

A
$$\dfrac{1}{\sqrt{2}}$$

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

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

D
None of these

Solution
Question 7
1 / -0

Let $$b = 4i + 3j $$ and $$c$$ be two vectors perpendicular to each other in the xy-plane. If $$ r_i, \ i =1, 2 ... n$$, are the vectors in the same plane having projections $$1 $$ and $$2$$ along $$b$$ and $$c$$ respectively then $$ \displaystyle \sum_{i=1}^{n} \left| r_{i}\right|^{2}$$ is equal to

A
$$20$$

B
$$10$$

C
$$4$$

D
$$7$$

Solution

Let $$ r=\lambda b + \mu c $$ and $$c = \pm (xi + yj).$$ Since $$c$$ and $$b$$ are perpendicular, we have
$$\displaystyle 4x + 3y = 0 \Rightarrow c= \pm x\left( i-\frac{4}{3}j\right)$$
$$ \pm 1= proj\cdot of $$r$$ on $$b$$ \displaystyle=\frac{ r\cdot b}{\left| b\right| }=\frac{(\lambda b+\mu c)\cdot b}{\left| b\right| }
=\frac{\lambda b\cdot b}{\left| b\right| } [\because b\cdot c=0]$$
$$ \lambda \left| b\right| =5 \lambda $$ Hence $$ \lambda=\pm 1/5$$
Also $$ \pm 2=proj.$$ of $$ r $$ on $$\displaystyle c=\frac {r\cdot c}{\left|c\right|}$$
$$\displaystyle =\frac{(\lambda b+\mu c)\cdot c}{\left| c\right|} =\mu \left| c\right| =\frac{5}{3} \mu x$$
Thus, $$ \mu x =\pm 6/5$$ Therefore,
$$\displaystyle r=\frac{1}{5} (4i+3j)+\frac{6}{5}\left( i-\frac{4}{3}j\right) =\pm (2i-j,)$$
$$\displaystyle r=\frac{1}{5} (4i+3j)-\frac{6}{5}\left( i-\frac{4}{3}j\right) =\pm \left (-\frac{2}{5} i+ \frac{11}{5}j,\right)$$
Thus there are four such vectors
$$ \displaystyle \sum_{i=1}^{4} \left| r_{i}\right|^{2}=2 \left| 2i -j\right| ^{2} +\left| -\frac{ 2}{5} i + \frac{11}{5} j\right|^{2} =20$$
Question 8
1 / -0

The magnitude of the projection of the vector $$\overline{a} =4\overline{i}-3\overline{j}+2\overline{k}$$ on the line which makes equal angles with the coordinate axes is

A
$$\sqrt{2}$$

B
$$\sqrt{3}$$

C
$$\dfrac{1}{\sqrt{3}}$$

D
$$\dfrac{1}{\sqrt{2}}$$

Solution

Given $$a=4i-3j+2k$$

The line which makes equal angle with the coordinate axis is $$b=i+j+k$$

The projection of $$a$$ on $$b$$ is $$\dfrac{a.b}{b} = \dfrac{4-3+2}{\sqrt3}=\dfrac{3}{\sqrt3}=\sqrt3$$
Question 9
1 / -0

Let $$ x_{0}$$ and $$x_{1}$$ be the critical points of $$\displaystyle f(x) =\int_{1}^{x}(t(t + 1) (t + 2) (t + 3)- 24).dt $$ and $$ \vec r$$ & $$\vec r'$$ be the parallel vectors with $$\left| \vec r \right| =\left| x_{0}\right| $$ and $$\left| \vec r\ ' \right| =\left|x_{1}\right| ,$$ then $$ \vec r\cdot \vec r\ '$$ is equal to

A
$$24$$

B
$$16$$

C
$$8$$

D
$$4$$

Solution

$$\displaystyle f(x)=\int _{ 1 }^{ x } (t(t+1)(t+2)(t+3)24)dt$$
$$ f'( x) =x(x + 1)(x + 2)(x + 3) -24 $$,
Clearly $$ f'(1) =0,$$
Dividing $$ f'(x)$$ by $$ x -1$$, we get
$$ f'(x) = (x -1)(x + 4)(x^{2} + 3x + 6)$$
Thus the critical points are $$ x = 1, -4.$$
Hence $$ r \cdot r'= \left| x_{0}\right| \left| x_{1}\right| =1\times 4 = 4.$$
Question 10
1 / -0

If one point on the vector $$2i -4j-k$$ is $$(2,1,3)$$,the other point is?

A
$$(-4,3,2)$$

B
$$(4,-3,-2)$$

C
$$(3,2,1)$$

D
$$(4,-3,2)$$

Solution

Let the other point on the vector $$2i-4j-k$$ be $$(x,y,z)$$. Given point on the vector is $$(2,1,3)$$.

Thus the vector is represented as
$$(2,-4,-1) = (x,y,z)-(2,1,3)$$
$$(2,-4,-1)=(x-2,y-1,z-3)$$

Equating the corresponding points we get $$2=x-2, -4=y-1, -1=z-3$$

$$\Rightarrow x=4, y=-3, z=2$$

Hence the other point on the vector is $$(4,-3,2)$$.

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 68

Result

Vector Algebra Test - 68

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SHARING IS CARING

Question 1

$$2 \widehat{i} + 3 \widehat{j} - 3 \widehat{k}$$

$$- 2 \widehat{i} - \widehat{j} + 5 \widehat{k}$$

$$2 \widehat{i} + 3 \widehat{j} + 3 \widehat{k}$$

$$2 \widehat{i} + \widehat{j} + 5 \widehat{k}$$

Solution

Question 2

Incentre

Circumcentre

Centroid

Orthocentre

Solution

Question 3

$$(\vec{a} \times \vec{c}) \times (\vec{b} \times \vec{d}) = \vec{0}$$

$$(\vec{a} \times \vec{c}) \cdot (\vec{b} \times \vec{d}) = \vec{0}$$

$$(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0}$$

$$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = \vec{0}$$

Solution

Question 4

$$\;2\sqrt{5}$$

$$\;2\sqrt{2}$$

$$\;10\sqrt{5}$$

$$\;5\sqrt{2}$$

Solution

Question 5

$$2\overrightarrow{AO}$$

$$4\overrightarrow{AO}$$

$$6\overrightarrow{AO}$$

Zero

Solution

Question 6

$$\dfrac{1}{\sqrt{2}}$$

$$\dfrac{1}{\sqrt{3}}$$

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

None of these

Solution

Question 7

$$20$$

$$10$$

$$4$$

$$7$$

Solution

Question 8

$$\sqrt{2}$$

$$\sqrt{3}$$

$$\dfrac{1}{\sqrt{3}}$$

$$\dfrac{1}{\sqrt{2}}$$

Solution

Question 9

$$24$$

$$16$$

$$8$$

$$4$$

Solution

Question 10

$$(-4,3,2)$$

$$(4,-3,-2)$$

$$(3,2,1)$$

$$(4,-3,2)$$

Solution

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