# If $a=2\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,-m\overset{\wedge }{\mathop{k}}\,\text{ and }b=\dfrac{4}{7}\overset{\wedge }{\mathop{i}}\,-\dfrac{2}{7}\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\,$ are collinear, then the value of m is equal to

(A) -7

(B) -1

(C) 2

(D) 7

(E) -2

Answer

Verified

381.6k+ views

Hint: Two vectors are collinear if they are in the same line. Use formula $\overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{B}}\,=\left( \overset{\to }{\mathop{A}}\, \right)\left( \overset{\to }{\mathop{B}}\, \right)\cos \theta $ where $\theta $is angle between $\overset{\to }{\mathop{A}}\,\text{ and }\overset{\to }{\mathop{B}}\,$.

Complete step-by-step answer:

Here, we have two vectors given as $\overset{\to }{\mathop{a}}\,=2i-j-mk\text{ and }\overset{\to }{\mathop{b}}\,=\dfrac{4}{7}i-\dfrac{2}{7}j+2k.$

Now, we have to find value of m if $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$ are collinear.

As, we know collinear means in a line, that is if two vectors $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$ are collinear then they lie in the same line.

In another language, we can say that if two vectors are collinear then the angle between them is $0{}^\circ $.

Hence, $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$ have angle $0{}^\circ $ between them.

Now, we know dot product of two vectors can be given by;

$\overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{B}}\,=\left( \overset{\to }{\mathop{A}}\, \right)\left( \overset{\to }{\mathop{B}}\, \right)\cos \theta $

Where $\theta $ is angle between $\overset{\to }{\mathop{A}}\,\text{ and }\overset{\to }{\mathop{B}}\,$.

As, we have two vectors $\overset{\to }{\mathop{a}}\,=2i-j-mk\text{ and }\overset{\to }{\mathop{b}}\,=\dfrac{4}{7}i-\dfrac{2}{7}j+2\overset{\wedge }{\mathop{k}}\,$ which have $0{}^\circ $ between them.

Hence,

$\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|\cos \theta $………………….(1)

We know magnitude of any vector $\overset{\to }{\mathop{r}}\,=xi-yj-zk$ is given as;

$\left| \overset{\to }{\mathop{r}}\, \right|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$

Hence, equation (1) can be written as

We know that $\cos 0{}^\circ =1$ and can simplify the above equation as;

$\begin{align}

& \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=\sqrt{5+{{m}^{2}}}\sqrt{\dfrac{20}{49}+4}\left( 1 \right) \\

& \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=\sqrt{5+{{m}^{2}}}\dfrac{\sqrt{216}}{7} \\

& or \\

& \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=\dfrac{6\sqrt{6}}{7}\sqrt{5+{{m}^{2}}}.............\left( 2 \right) \\

\end{align}$

Now, we can rewrite $\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,$ in different way as;

$\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=\left( 2i-j-mk \right).\left( \dfrac{4}{7}i-\dfrac{2}{7}j+2k \right)............\left( 3 \right)$

As we know the same vectors have angle $0{}^\circ $. Hence, angles between (i, j), (j, k) and (i, k) are perpendicular to each other as they are unit vectors through the x, y and z axis.

Hence, equation (3) can be solved as;

$\begin{align}

& \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=2\times \dfrac{4}{7}+1\times \dfrac{2}{7}-m\times 2 \\

& \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=\dfrac{10}{7}-2m..............\left( 4 \right) \\

\end{align}$

Now, equation (2) and (4) have equal L.H.S hence R.H.S should also be equal to each other.

$\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=\dfrac{6\sqrt{6}}{7}\sqrt{5+{{m}^{2}}}=\dfrac{10}{7}-2m$

Squaring both sides of the equation we get;

$\begin{align}

& \dfrac{216\left( 5+{{m}^{2}} \right)}{49}=\dfrac{{{\left( 10-14m \right)}^{2}}}{49} \\

& 216\times 5+216{{m}^{2}}=100+196{{m}^{2}}-280m \\

& 20{{m}^{2}}+280m+980=0 \\

\end{align}$

Dividing the whole equation by 20, we get;

${{m}^{2}}+14m+49=0$

Now, we can factorize above equation as;

${{\left( m+7 \right)}^{2}}=0$

Hence,

$\begin{align}

& m+7=0 \\

& m=-7 \\

\end{align}$

Therefore, for m = -7, $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$ as given in question are collinear.

Note: One can go wrong while putting angle between two vectors $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$which is $0{}^\circ $. He/she may put the angle be $180{}^\circ $ which is wrong.

Need to be careful with angles between $\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j\ }}\,$and $\overset{\wedge }{\mathop{k}}\,$ during evaluation of $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$.

Complete step-by-step answer:

Here, we have two vectors given as $\overset{\to }{\mathop{a}}\,=2i-j-mk\text{ and }\overset{\to }{\mathop{b}}\,=\dfrac{4}{7}i-\dfrac{2}{7}j+2k.$

Now, we have to find value of m if $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$ are collinear.

As, we know collinear means in a line, that is if two vectors $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$ are collinear then they lie in the same line.

In another language, we can say that if two vectors are collinear then the angle between them is $0{}^\circ $.

Hence, $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$ have angle $0{}^\circ $ between them.

Now, we know dot product of two vectors can be given by;

$\overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{B}}\,=\left( \overset{\to }{\mathop{A}}\, \right)\left( \overset{\to }{\mathop{B}}\, \right)\cos \theta $

Where $\theta $ is angle between $\overset{\to }{\mathop{A}}\,\text{ and }\overset{\to }{\mathop{B}}\,$.

As, we have two vectors $\overset{\to }{\mathop{a}}\,=2i-j-mk\text{ and }\overset{\to }{\mathop{b}}\,=\dfrac{4}{7}i-\dfrac{2}{7}j+2\overset{\wedge }{\mathop{k}}\,$ which have $0{}^\circ $ between them.

Hence,

$\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|\cos \theta $………………….(1)

We know magnitude of any vector $\overset{\to }{\mathop{r}}\,=xi-yj-zk$ is given as;

$\left| \overset{\to }{\mathop{r}}\, \right|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$

Hence, equation (1) can be written as

We know that $\cos 0{}^\circ =1$ and can simplify the above equation as;

$\begin{align}

& \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=\sqrt{5+{{m}^{2}}}\sqrt{\dfrac{20}{49}+4}\left( 1 \right) \\

& \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=\sqrt{5+{{m}^{2}}}\dfrac{\sqrt{216}}{7} \\

& or \\

& \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=\dfrac{6\sqrt{6}}{7}\sqrt{5+{{m}^{2}}}.............\left( 2 \right) \\

\end{align}$

Now, we can rewrite $\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,$ in different way as;

$\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=\left( 2i-j-mk \right).\left( \dfrac{4}{7}i-\dfrac{2}{7}j+2k \right)............\left( 3 \right)$

As we know the same vectors have angle $0{}^\circ $. Hence, angles between (i, j), (j, k) and (i, k) are perpendicular to each other as they are unit vectors through the x, y and z axis.

Hence, equation (3) can be solved as;

$\begin{align}

& \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=2\times \dfrac{4}{7}+1\times \dfrac{2}{7}-m\times 2 \\

& \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=\dfrac{10}{7}-2m..............\left( 4 \right) \\

\end{align}$

Now, equation (2) and (4) have equal L.H.S hence R.H.S should also be equal to each other.

$\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=\dfrac{6\sqrt{6}}{7}\sqrt{5+{{m}^{2}}}=\dfrac{10}{7}-2m$

Squaring both sides of the equation we get;

$\begin{align}

& \dfrac{216\left( 5+{{m}^{2}} \right)}{49}=\dfrac{{{\left( 10-14m \right)}^{2}}}{49} \\

& 216\times 5+216{{m}^{2}}=100+196{{m}^{2}}-280m \\

& 20{{m}^{2}}+280m+980=0 \\

\end{align}$

Dividing the whole equation by 20, we get;

${{m}^{2}}+14m+49=0$

Now, we can factorize above equation as;

${{\left( m+7 \right)}^{2}}=0$

Hence,

$\begin{align}

& m+7=0 \\

& m=-7 \\

\end{align}$

Therefore, for m = -7, $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$ as given in question are collinear.

Note: One can go wrong while putting angle between two vectors $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$which is $0{}^\circ $. He/she may put the angle be $180{}^\circ $ which is wrong.

Need to be careful with angles between $\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j\ }}\,$and $\overset{\wedge }{\mathop{k}}\,$ during evaluation of $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$.

Recently Updated Pages

Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Find the values of other five trigonometric ratios class 10 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts

The lightest gas is A nitrogen B helium C oxygen D class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Which place is known as the tea garden of India class 8 social science CBSE

What is pollution? How many types of pollution? Define it

Write a letter to the principal requesting him to grant class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE