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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 $\overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,$.

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