Vector Algebra Test

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Vector Algebra Test - 40

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Question 1
1 / -0

$$AB=3i+j-k$$ and $$AC=i-j+3k$$. If the point $$P$$ on the line segment $$BC$$ is equidistant from $$AB$$ and $$AC,$$ then $$AP$$ is

A
$$2i-k$$

B
$$i-2k$$

C
$$2i+k$$

D
None of these

Solution

A point equidistant from $$AB$$ and $$AC$$ is on the bisector of the angle $$BAC.$$
A vector along the internal bisector of the angle $$BAC$$
$$\displaystyle=\frac { AB }{ \left| AB \right| } +\frac { AC }{ \left| AC \right| } $$
$$\displaystyle=\frac { 3i+j-k }{ \sqrt { 9+1+1 } } +\frac { i-j+3k }{ \sqrt { 1+1+9 } } =\frac { 1 }{ \sqrt { 11 } } \left( 4i+2k \right) \\ \therefore AP=t\left( 2i+k \right) \\ \therefore BP=AP-AB=t\left( 2i+k \right) -\left( 3i+j-k \right) =\left( 2t-3 \right) i-j+\left( t+1 \right) k$$
Also $$BC=AC-AB=\left( i-j+3k \right) -\left( 3i+j-k \right) =-2i-2j+4k$$
But $$BP=sBC$$
$$\therefore \left( 2t-3 \right) i-j+\left( t+1 \right) k=s\left( -2i-2j+4k \right) \\ \therefore 2t-3=-2s,-1=-2s,t+1=4s$$
$$\displaystyle\therefore s=\frac { 1 }{ 2 } $$ and $$t=1$$
$$\therefore AP=2i+k$$
Question 2
1 / -0

If $$4i+7j+8k, 2i+7j+7k$$ and $$3i+5j+7k$$ are the position vectors of the vertices $$A,B$$ and $$C$$ respectively of triangle $$ABC$$. The position vector of the point where the bisector of angle $$A$$ meets $$BC.$$

A
$$\displaystyle \frac { 1 }{ 3 } \left( 5j+12k \right) $$

B
$$\displaystyle \frac { 1 }{ 3 } \left( 6i+13j+18k \right) $$

C
$$\displaystyle \frac { 2 }{ 3 } \left( 6i+8j+6k \right) $$

D
$$\displaystyle \frac { 2 }{ 3 } \left( -6i-8j-6k \right) $$

Solution

Suppose the bisector of angle $$A$$ meets $$BC$$ at $$D$$.
Then, $$AD$$ divides $$BC$$ in the ratio $$AB:AC$$
$$\therefore$$ position vector of $$D$$ $$\displaystyle =\frac { \left( AC \right) \left( 2i+3j+4k \right)+ \left( AB \right) \left( 2i+5j+7k \right) }{ \left( AB \right) +\left( AC \right) } $$
But $$AB+-2i-4j-4k$$ and $$AC=-2i-2j-k$$
$$\therefore AB= \sqrt{4+16+11}=6$$ and $$AC=\sqrt{4+4+1}=3$$
$$\therefore$$ P.V. of $$D$$ is $$\displaystyle \frac { 6\left( 2i+5j+7k \right) +3\left( 2i+3j+4k \right) }{ 6+3 } $$
$$\displaystyle =\frac { 18i+39j+54k }{ 9 } =\frac { 1 }{ 3 } \left( 6i+13j+18k \right) $$
Question 3
1 / -0

A straight line $$'L'$$ cuts the sides $$AB,AC$$ and $$AD$$ of a parallelogram $$ABCD$$ at points $${ B }_{ 1 },{ C }_{ 1 }$$ and $${D}_{1}$$ respectively. If $$\overrightarrow { A{ B }_{ 1 } } ={ \lambda }_{ 1 }\overrightarrow { AB } ,\overrightarrow { A{ D }_{ 1 } } ={ \lambda }_{ 2 }\overrightarrow { AD } $$ and $$\overrightarrow { A{ C }_{ 1 } } ={ \lambda }_{ 3 }\overrightarrow { AC } $$, then $$\displaystyle \frac { 1 }{ { \lambda }_{ 3 } } $$ is equal to

A
$$\displaystyle \frac { 1 }{ { \lambda }_{ 1 } } +\frac { 1 }{ { \lambda }_{ 2 } } $$

B
$$\displaystyle \frac { 1 }{ { \lambda }_{ 1 } } -\frac { 1 }{ { \lambda }_{ 2 } } $$

C
$$-{ \lambda }_{ 1 }+{ \lambda }_{ 2 }$$

D
$${ \lambda }_{ 1 }-{ \lambda }_{ 2 }$$

Solution

Let $$\displaystyle \overrightarrow { AB } =\overrightarrow { a } ,\overrightarrow { AD } =\overrightarrow { b } ,$$
then $$\displaystyle \overrightarrow { AC } =\overrightarrow { a } +\overrightarrow { b } $$
Given $$\displaystyle \overrightarrow { A{ B }_{ 1 } } ={ \lambda }_{ 2 }\overrightarrow { a } ,\overrightarrow { A{ D }_{ 1 } } ={ \lambda }_{ 2 }\overrightarrow { b } ,\overrightarrow { A{ C }_{ 1 } } ={ \lambda }_{ 3 }\left( \overrightarrow { a } +\overrightarrow { b } \right) $$
$$\displaystyle \overrightarrow { { B }_{ 1 }{ D }_{ 1 } } =\overrightarrow { { AD }_{ 1 } } -\overrightarrow { { AB }_{ 1 } } ={ \lambda }_{ 2 }\overrightarrow { b } -{ \lambda }_{ 2 }\overrightarrow { b } -{ \lambda }_{ 1 }\overrightarrow { a } $$

Since vectors $$\displaystyle \overrightarrow { { D }_{ 1 }{ C }_{ 1 } } $$ and
$$\displaystyle \overrightarrow { { B }_{ 1 }{ D }_{ 1 } } $$ are collinear, we have
$$\displaystyle \overrightarrow { { D }_{ 1 }{ C }_{ 1 } } =k\overrightarrow { { B }_{ 1 }{ D }_{ 1 } } $$ for some $$\displaystyle k\in R$$
$$\displaystyle \Rightarrow \overrightarrow { { AC }_{ 1 } } -\overrightarrow { { AD }_{ 1 } } =k\overrightarrow { { B }_{ 1 }{ D }_{ 1 } } $$
$$\displaystyle \Rightarrow { \lambda }_{ 3 }\left( \overrightarrow { a } +\overrightarrow { b } \right) -{ \lambda }_{ 2 }\overrightarrow { b } =k.\left( { \lambda }_{ 2 }\overrightarrow { b } -{ \lambda }_{ 1 }\overrightarrow { a } \right) $$
$$\displaystyle \Rightarrow { \lambda }_{ 3 }\overrightarrow { a } +\left( { \lambda }_{ 3 }-{ \lambda }_{ 2 } \right) \overrightarrow { b } =k.{ \lambda }_{ 2 }\overrightarrow { b } -k.{ \lambda }_{ 1 }\overrightarrow { a } $$

Hence, $$\displaystyle { \lambda }_{ 3 }=-k{ \lambda }_{ 1 }$$ and $$\displaystyle { \lambda }_{ 3 }-{ \lambda }_{ 2 }=k{ \lambda }_{ 2 }$$
$$\displaystyle \Rightarrow k=-\frac { { \lambda }_{ 3 } }{ { \lambda }_{ 1 } } =\frac { { \lambda }_{ 3 }-{ \lambda }_{ 2 } }{ { \lambda }_{ 2 } }$$
$$\Rightarrow { \lambda }_{ 1 }{ \lambda }_{ 2 }={ \lambda }_{ 1 }{ \lambda }_{ 3 }+{ \lambda }_{ 2 }{ \lambda }_{ 3 }$$
$$\displaystyle \Rightarrow \frac { 1 }{ { \lambda }_{ 3 } } =\frac { 1 }{ { \lambda }_{ 1 } } +\frac { 1 }{ { \lambda }_{ 2 } } $$
Question 4
1 / -0

The sum of the three vectors determined by the medians of a triangle directed from the vertices is

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$\displaystyle \frac{1}{3}$$

Solution

$$\because D$$ is middle point of $$BC$$
$$\displaystyle \therefore AD=\frac { AB+AC }{ 2 } $$ ...(1)
$$E$$ is middle point of $$CA$$
$$\displaystyle \therefore BE=\frac { BA+BC }{ 2 } $$ ....(2)
$$F$$ is middle point of $$AB$$
$$\displaystyle \therefore CF=\frac { CB+CA }{ 2 } $$ ...(3)
Adding (1),(2) and (3), we get
$$AD+BE+CF=0$$
Question 5
1 / -0

In a parallelogram $$ABCD$$, $$\left| AB \right| =a,\left| AD \right| =b$$ and $$\left| AC \right| =c$$. Then $$DB.AB$$ has the value

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

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

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

D
$$\displaystyle \frac { { a }^{ 2 }+3{ b }^{ 2 }+{ c }^{ 2 } }{ 2 } $$
Question 6
1 / -0

If $$O$$ and $$O'$$ are circumcenter and orthocenter of a triangle $$ABC$$ then $$\left( OA+OB+OC \right) $$ equals

A
$$2OO'$$

B
$$OO'$$

C
$$O'O$$

D
$$2O'O$$

Solution

according to the properties of triangle
circumcentre O in triangle ABC
the sum of distance from circumcentre to vertices of ABC is the distance from circumcentre to orthocentre
$$(OA+OB+OC)=OO'$$
Question 7
1 / -0

The position vectors of three consecutive vertices of a parallelogram are $$i+j+k, i+3j+5k$$ and $$7i+9j+11k$$. The position vector of the fourth vertex is

A
$$6(i+j+k)$$

B
$$7(i+j+k)$$

C
$$2j-4k$$

D
$$6i+8j+10k$$

Solution

Let $$A\left( \overrightarrow { a } \right) =i+j+k$$
$$B\left( \overrightarrow { b } \right) =i+3j+5k\\ C\left( \overrightarrow { c } \right) =7i+9j+11k$$
$$D\left( \overrightarrow { d } \right) $$ is to be determined
$$ABCD$$ is a parallelogram so $$\displaystyle \frac { 1 }{ 2 } \left( \overrightarrow { b } +\overrightarrow { d } \right) =\frac { 1 }{ 2 } \left( \overrightarrow { a } +\overrightarrow { c } \right) $$
$$\overrightarrow { d } =\overrightarrow { a } +\overrightarrow { c } -\overrightarrow { b } =8i+10j+12k-\left( i+3j+5k \right) \\ =7i+7j+7k=7\left( i+j+k \right) $$
Question 8
1 / -0

Let $$G$$ be the centroid of a triangle $$ABC$$. If $$AB=a,AC=b$$ then the bisector $$AG$$, in terms of vectors $$a$$ and $$b$$ is

A
$$\displaystyle \frac { 2 }{ 3 } \left( a+b \right) $$

B
$$\displaystyle \frac { 1 }{ 6 } \left( a+b \right) $$

C
$$\displaystyle \frac { 1 }{ 3 } \left( a+b \right) $$

D
$$\displaystyle \frac { 1 }{ 2 } \left( a+b \right) $$

Solution

Take $$A$$ as origin.
Then position vectors of $$A,B,C$$ are given $$0,a$$ and $$b$$ respectively.
Then the centroid $$G$$ has position vector
$$\displaystyle \frac { 0+a+b }{ 3 } =\frac { a+b }{ 3 } $$
$$\displaystyle \therefore AG=\frac { a+b }{ 3 } $$
Question 9
1 / -0

The value of $$\vec i \times (\vec a \times \vec i) + \vec j \times (\vec a \times \vec j) + \vec k \times (\vec a \times \vec k)$$ is (where $$\vec i, \vec j, \vec k$$ are unit vectors)

A
$$\vec a$$

B
$$2 \vec a$$

C
0

D
$$- \vec a$$

Solution

$$i \times (\bar a \times i) = (i . i) \bar a - (i. \bar a) i = \bar a - (i . \bar a)i$$
so $$i \times (\bar a \times i) + j \times (\bar a \times j) + k \times (\bar a \times k)$$
$$= \bar a - (i . \bar a) i + \bar - (j . \bar a) j + \bar a - (k . \bar a) k$$
$$= 3\bar a - [(i . \bar a) i + (j . \bar a) j + (k . \bar a) k]$$
$$= 3 \bar a - \bar a = 2 \bar a$$.
Question 10
1 / -0

Let $$\vec{a}=\hat{i}+\hat{j}+3\hat{k}\;\&\;\vec{b}=2\hat{i}-3\hat{j}+4\hat{k}$$. If projection of $$\vec{a}$$ on $$\vec{b}$$ is $$\displaystyle\frac{k}{\sqrt{29}}$$, then the value of $$(k-2)$$ is

A
$$9$$

B
$$-9$$

C
$$8$$

D
$$6$$

Solution

GIVEN
$$\vec{a}=\hat{i}+\hat{j}+3\hat{k}$$
$$\vec{b}=2\hat{i}-3\hat{j}+4\hat{k}$$
$$Proj_{a}b=\dfrac{\vec{a}\cdot\vec{b}}{\left | \vec{b} \right |}$$
$$Proj_{a}b=\dfrac{(\hat{i}+\hat{j}+3\hat{k})\cdot(2\hat{i}-3\hat{j}+4\hat{k})}{\left | \sqrt{2^2+(-3)^2+4^2} \right |}$$
$$Proj_{a}b=\dfrac{2-3+12}{\sqrt{4+9+16}}$$
$$Proj_{a}b=\dfrac{11}{ \sqrt{29}}$$
comparing RHS with$$\dfrac{k}{ \sqrt{29}}$$
$$k=11$$
$$k-2=11-2$$
$$k-2=9$$

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 40

Result

Vector Algebra Test - 40

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SHARING IS CARING

Question 1

$$2i-k$$

$$i-2k$$

$$2i+k$$

None of these

Solution

Question 2

$$\displaystyle \frac { 1 }{ 3 } \left( 5j+12k \right) $$

$$\displaystyle \frac { 1 }{ 3 } \left( 6i+13j+18k \right) $$

$$\displaystyle \frac { 2 }{ 3 } \left( 6i+8j+6k \right) $$

$$\displaystyle \frac { 2 }{ 3 } \left( -6i-8j-6k \right) $$

Solution

Question 3

$$\displaystyle \frac { 1 }{ { \lambda }_{ 1 } } +\frac { 1 }{ { \lambda }_{ 2 } } $$

$$\displaystyle \frac { 1 }{ { \lambda }_{ 1 } } -\frac { 1 }{ { \lambda }_{ 2 } } $$

$$-{ \lambda }_{ 1 }+{ \lambda }_{ 2 }$$

$${ \lambda }_{ 1 }-{ \lambda }_{ 2 }$$

Solution

Question 4

$$0$$

$$1$$

$$-1$$

$$\displaystyle \frac{1}{3}$$

Solution

Question 5

$$\displaystyle \frac { 3{ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2 } $$

$$\displaystyle \frac { { a }^{ 2 }+3{ b }^{ 2 }-{ c }^{ 2 } }{ 2 } $$

$$\displaystyle \frac { { a }^{ 2 }-{ b }^{ 2 }+3{ c }^{ 2 } }{ 2 } $$

$$\displaystyle \frac { { a }^{ 2 }+3{ b }^{ 2 }+{ c }^{ 2 } }{ 2 } $$

Question 6

$$2OO'$$

$$OO'$$

$$O'O$$

$$2O'O$$

Solution

Question 7

$$6(i+j+k)$$

$$7(i+j+k)$$

$$2j-4k$$

$$6i+8j+10k$$

Solution

Question 8

$$\displaystyle \frac { 2 }{ 3 } \left( a+b \right) $$

$$\displaystyle \frac { 1 }{ 6 } \left( a+b \right) $$

$$\displaystyle \frac { 1 }{ 3 } \left( a+b \right) $$

$$\displaystyle \frac { 1 }{ 2 } \left( a+b \right) $$

Solution

Question 9

$$\vec a$$

$$2 \vec a$$

0

$$- \vec a$$

Solution

Question 10

$$9$$

$$-9$$

$$8$$

$$6$$

Solution

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