Vector Algebra Test

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Vector Algebra Test - 26

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Question 1
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If the vertices of a $$\Delta ABC$$ are $$A= (1,-1,-3)$$ , $$B= (2, 1, -2)$$ and $$C=(-5,2,-6)$$ then the length of the internal bisector of angle $$A$$ is

A
$$\displaystyle \dfrac{3\sqrt{10}}{2}$$

B
$$\displaystyle \dfrac{3\sqrt{10}}{5}$$

C
$$\displaystyle \dfrac{3\sqrt{10}}{7}$$

D
$$\displaystyle \dfrac{3\sqrt{10}}{4}$$

Solution

$$AD$$ bisects the $$BC$$ in ratio of $$\dfrac{|AB|}{|AC|}=\dfrac{|BD|}{|CD|}=\dfrac{\sqrt6}{\sqrt{54}}=\dfrac{1}{3}$$

$$\dfrac{BD}{DC}=\dfrac{1}{3}=\dfrac{c}{b}$$

Coordinates of $$D$$ is $$=\dfrac{1}{4}(3B+C)$$ (section formula)

$$D(\dfrac{1}{4},\dfrac{5}{4},-3)$$

Length of $$ AD=\sqrt{(1-\dfrac{1}{4})^{2}+(-1-\dfrac{5}{4})^{2}+(-3+3)^{2}}$$

$$=\dfrac{\sqrt{90}}{4}$$

$$=\dfrac{3\sqrt{10}}{4}$$
Question 2
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The condition for the vectors $$\vec{a},\vec{b},\vec{c},\vec{d}$$ to be the sides of a parallelogram taken in order is

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

B
$$\vec{a}+\vec{b}=\vec{c}+\vec{d}=0$$

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

D
$$\vec{a}+\vec{c}=\vec{b}+\vec{d}=0$$

Solution

Consider the vectors $$\vec{a}$$ and $$\vec{b}$$.
Now, the sides with length as $$\vec a , \vec b$$ forms two sides of a triangle and diagonal forms the third side of the same triangle.
$$\therefore$$ By using triangle law of vector addition, the diagonal is given by
$$\vec{a}+\vec{b}$$ ........... (i)
Now consider the vectors $$\vec{c}$$ and $$\vec{d}$$.
The same diagonal can be written as $$\vec{c}+\vec{d}$$ using triangle law of vector addition.
$$\therefore$$ $$\vec{a}+\vec{b}=\vec{c}+\vec{d}$$.
Hence, option A is correct.
Question 3
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Let $$A$$ and $$B$$ be points with position vectors $$\overline{a}$$ and $$\overline{b}$$ with respect to origin $$O$$. If the point $$C$$ on $$OA$$ is such that $$2\overline{AC}=\overline{CO}$$, $$\overline{CD}$$ is parallel to $$\overline{OB}$$ and $$|\overline{CD}|=3|\overline{OB}|$$ then $$AD$$ is

A
$$\displaystyle \overline{b}-\dfrac{a}{9}$$

B
$$3\displaystyle \overline{b}-\dfrac{a}{3}$$

C
$$\displaystyle \overline{b}-\dfrac{a}{3}$$

D
$$\displaystyle \overline{b}+\dfrac{a}{3}$$

Solution

Given that $$2 \overrightarrow{AC}=\overrightarrow{CO}$$
$$2(\vec{c}-\vec{a})=-\vec{c}$$
$$3\vec{c}=2\vec{a}$$
$$\vec{c}=\dfrac{2\vec{a}}{3}$$
Also given $$\overrightarrow{CD}\left | \right |\overrightarrow{OB}$$
$$\Rightarrow \vec{a}-\vec{c}=k\vec{b}$$
$$\Rightarrow \left | \vec{d}-\vec{c} \right |=3\left | \vec{b} \right |$$
$$\Rightarrow k\left | \vec{b} \right |=3\left | \vec{b} \right |$$
$$\Rightarrow k=3$$
and $$\vec{d}=\vec{c}+3\vec{b}$$
$$\Rightarrow \overrightarrow{AD}=\vec{d}-\vec{a}$$ $$=3\vec{b}+\vec{c}-\vec{a}$$
$$\Rightarrow \overrightarrow{AD}=3\vec{b}-\dfrac{\vec{a}}{3}$$
Question 4
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The adjacent sides of a parallelogram are $$2{ \hat { i } }+4{ \hat { j } }-5{ \hat { k } }$$ and $$ { \hat { i } }+2{ \hat { j } }{ + }3{ \hat { k } }$$ then the unit vector parallel to a diagonal is

A
$$\displaystyle \dfrac{-\overline{i}-2\overline{j}+8\overline{k}}{\sqrt{69}}$$

B
$$\displaystyle \dfrac{3\overline{i}+6\overline{j}-2\overline{k}}{7}$$

C
$$\displaystyle \dfrac{\overline{i}+2\overline{j}-8\overline{k}}{\sqrt{69}}$$

D
All the above

Solution

As in this question direction of vector is not specified then it can be
$$\pm \vec{a}\ \pm \vec{b}=$$ Diagonal Vectors
Diagonal vector =$$\dfrac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\dfrac{3i+6\hat{j}-2\hat{k}}{\sqrt{49}}$$
or $$\dfrac{\vec{a}-\vec{b}}{|\vec{a}-\vec{b}|}=\dfrac{i+2\hat{j}-8\hat{k}}{\sqrt{69}}$$

or $$\dfrac{\vec{-a}+\vec{b}}{|\vec{-a}+\vec{b}|}=\dfrac{-i-2\hat{j}+8\hat{k}}{\sqrt{69}}$$
Question 5
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If the diagonals of a parallelogram are $$\overline{i}+5\overline{j}-2\overline{k}$$ and $$-2\overline{i}+\overline{j}+3\overline{k}$$, then the lengths of its sides are

A
$$\displaystyle \dfrac{\sqrt{38}}{2},\ \displaystyle \dfrac{\sqrt{50}}{2}$$

B
$$\displaystyle \dfrac{\sqrt{35}}{2},\dfrac{\sqrt{45}}{2}$$

C
$$\sqrt{38},\ \sqrt{50}$$

D
$$\sqrt{35},\ \sqrt{45}$$

Solution

Let $$A$$ be origin
$$\left | \overrightarrow{AC} \right |= (\hat{i}+5\hat{j}-2\hat{k})$$
$$(C=1,5,-2)$$
$$O$$ is mid point of $$AC$$, $$O= (\frac{1}{2}, \frac{5}{2}, -1)$$
$$B= (a,b,c)$$
$$\overrightarrow{OB}= \dfrac{1}{2}(\overrightarrow{BD})$$
$$(a\dfrac{-1}{2})\hat{i}+(b\dfrac{-5}{2})\hat{j}+(c+1)\hat{k}= -\hat{i}+\dfrac{1}{2}\hat{j}+\dfrac{3}{2}\hat{k}$$
$$a=\dfrac{-1}{2}\ b=3\ c= \dfrac{1}{2}$$
$$\left | \overrightarrow{AB} \right |= \sqrt{\frac{1}{4}+9+\dfrac{1}{4}}= \sqrt{\dfrac{38}{2}}$$
$$\left | \overrightarrow{BC} \right |= \sqrt{\frac{9}{4}+4+\dfrac{25}{4}}= \sqrt{\dfrac{50}{2}}$$
Question 6
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Let $$A({\vec{a}})$$ , $$B({\vec{b}}), C({\vec{c}})$$ be the vertices of the triangle $$ABC$$ and let $$DEF$$ 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
$$0$$

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

C
$$\displaystyle \dfrac{\vec{{a}}+\vec{{b}}+\vec{{c}}}{3}$$

D
$$\displaystyle \dfrac{2{\vec{a}}+{\vec{b}}+{\vec{c}}}{3}$$

Solution

Given :
$$A(\vec{a}), B(\vec{b}), C(\vec{c})$$ are the vertices of the triangle $$ABC$$.
$$D$$ is the midpoint of $$BC$$
$$\implies D=\left(\dfrac{\vec{b}+\vec{c}}{2}\right)$$
$$E$$ is the midpoint of $$CA$$
$$\implies E=\left(\dfrac{\vec{a}+\vec{c}}{2}\right)$$
$$F$$ is the midpoint of $$AB$$
$$\implies F=\left(\dfrac{\vec{a}+\vec{b}}{2}\right)$$
Now, $$AD, BE$$ and $$CF$$ are the medians of triangle $$ABC$$
All three medians intersects at point $$P$$.
$$\because\ P$$ divides $$AD$$ in the ratio $$2:1$$ and other medians also intersect at $$P$$.
$$\implies P$$ divides all three medians in the ratio $$2:1$$
$$\implies P$$ is the centroid of the triangle $$ABC$$.
$$\implies P= \dfrac{\vec{a}+\vec{b}+\vec{c}}{3}$$ ...... By def$$^n$$ of centroid.
Hence, option C is correct.
Question 7
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If $$I$$ is the centre of a circle inscribed in a triangle $$ABC$$, then $$|\overline{BC}|\overline{IA}+|\overline{CA}|\overline{IB}+|\overline{AB}|\overline{IC}$$ is

A
$$\overline{0}$$

B
$$\overline{IA}+\overline{IB}+\overline{IC}$$

C
$$\displaystyle \frac{\overline{IA}+\overline{IB}+\overline{IC}}{3}$$

D
$$\displaystyle \frac{\overline{IA}+\overline{IB}+\overline{IC}}{2}$$

Solution

We know that the position vector of the incenter of a $$\triangle ABC$$ is given by $$\displaystyle \frac { BC.\overline { a } +CA.\overline { b } +AB.\overline { c } }{ BC+CA+AB } $$
where $$\displaystyle \overline { a } ,\overline { b } ,\overline { c } $$ are the affixes of the vertices of the triangle.
Therefore, in the given triangle $$ABC,$$ the affix of $$I$$ is given by
$$\displaystyle \dfrac { \left| \overline { BC } \right| \overline { IA } +\left| \overline { CA } \right| \overline { IB } +\left| \overline { AB } \right| \overline { IC } }{ \left| \overline { BC } \right| +\left| \overline { CA } \right| +\left| \overline { AB } \right| } $$
But it is the position vector of $$I$$ with respect to $$O.$$
$$\displaystyle \therefore \ \overline { O } =\dfrac { \left| \overline { BC } \right| \overline { IA } +\left| \overline { CA } \right| \overline { IB } +\left| \overline { AB } \right| \overline { IC } }{ \left| \overline { BC } \right| +\left| \overline { CA } \right| +\left| \overline { AB } \right| } $$
$$\displaystyle \Rightarrow \left| \overline { BC } \right| \overline { IA } +\left| \overline { CA } \right| \overline { IB } +\left| \overline { AB } \right| \overline { IC } =0$$
Question 8
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Let $$A=2\hat{i}+4\hat{j}-\hat{k}$$, $$B=4\hat{i}+5\hat{j}{+}\hat{k}$$. If the centroid $$G$$ of the triangle $$ABC$$ is $$3\hat{i}+5\hat{j}-\hat{k}$$, then the position vector of $$C$$ is

A
$$3\hat{i}-6\hat{j}{+}3\hat{k}$$

B
$$3\hat{i}-6\hat{j}-3\hat{k}$$

C
$$3\hat{i}-6\hat{j}{+}2\hat{k}$$

D
$$3\hat{i}+6\hat{j}-3\hat{k}$$

Solution

Centroid is given by
$$G= \dfrac{\vec{A}+\vec{B}+\vec{C}}{3}$$
Putting given values to find $$\vec{C}$$
$$\Rightarrow \displaystyle 3(3\hat{i}+5\hat{j}-k)= 2\hat{i}+4\hat{j}-\hat{k}+4\hat{i}+5\hat{j}+\hat{k}+\vec{C}$$
$$\Rightarrow \displaystyle 3\hat{i}+6\hat{j}-3\hat{k}=\vec{C}$$
Question 9
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If $$\vec{{r}} {\times} \vec{{a}}=\vec{{b}}{\times}\vec{{a}};\ \vec{{r}}{\times}\vec{{b}}=\vec{{a}}{\times}\vec{{b}};\ \vec{a}\neq 0,\vec{b}\neq 0,\vec{a}\neq\lambda\vec{b};\ \vec{a}$$ is not perpendicular to $$\vec{b}$$, then $$\vec{r}=$$

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

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

C
$$\vec{{a}}{\times}\vec{{b}}+\vec{{a}}$$

D
$$\vec{{a}}{\times}\vec{{b}}+\vec{{b}}$$

Solution

Given, $$\vec{r} \times \vec{a} = -(\vec{a} \times \vec{b})$$ and $$\vec{r} \times \vec{b} = \vec{a} \times \vec{b}$$
Thus, $$\vec{r} \times \vec{a} = -(\vec{r} \times \vec{b})$$
$$\Rightarrow \vec{r} \times (\vec{a} + \vec{b}) = 0$$
$$\Rightarrow \vec{r} = \lambda(\vec{a} + \vec{b})$$
Hence, option B.
Question 10
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The plane $$2x-3y+z+6=0$$ divides the line segment joining $$(2, 4, 16)$$ and $$(3, 5, -4)$$ in the ratio

A
$$ 4 :5 $$

B
$$ 4:7 $$

C
$$2:1$$

D
$$1:2$$

Solution

Let the line segment joining points $$P(2,4,16)$$ and $$Q(3,5,-4)$$ be divided by the given plane in the ratio $$k:1$$ at the point $$R.$$
Then co-ordinates of $$R$$ are $$\displaystyle \left( \frac { 3k+2 }{ k+1 } ,\frac { 5k+4 }{ k+1 } ,\frac { -4k+16 }{ k+1 } \right) $$
Since $$R$$ lies on the plane $$2x-3y+z+6=0,$$
We have, $$\displaystyle 2\left( \frac { 3k+2 }{ k+1 } \right) -3\left( \frac { 5k+4 }{ k+1 } \right) +\left( \frac { -4k+16 }{ k+1 } \right) +6=0$$
$$\displaystyle \Rightarrow 2\left( 3k+2 \right) -3\left( 5k+4 \right) +\left( -4k+16 \right) +6\left( k+1 \right) =0$$
$$\displaystyle \Rightarrow -7k+14=0\Rightarrow k=2$$
$$\therefore PQ$$ is divided by the given plane in the ratio $$k:1$$ i.e., $$2:1$$