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Questions:

Generic 3

5 x 3 = 15

If the three vectors \(\vec{a}, \vec{b} \text { and } \vec{c}\) are given as \(a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}, b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k} \text { and } c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}\) Then, show that \(\vec{a} \times(\vec{b}+\vec{c})=(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\).

If a vector \(\vec{r}\) has magnitude 14and direction ratios 2, 3 and - 6. Then, find the direction cosines and components of \(\vec{r}\)given that \(\vec{r}\) makes an acute angle with x-axis.

It is given that:\(\overset { \rightarrow }{ x } =\frac { \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } ,\overset { \rightarrow }{ y } \frac { \overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } and\ \overset { \rightarrow }{ z } =\frac { \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } \) where \(\overset { \rightarrow }{ a } ,\overset { \rightarrow }{ b } ,\overset { \rightarrow }{ c } \) are non-coplanar vectors.

If \(\overrightarrow { a } \times \overrightarrow { b } =\overrightarrow { a } \times \overrightarrow { c } \ and\ \overrightarrow { a } \times \overrightarrow { c } =\overrightarrow { b } \times \overrightarrow { d } \) prove that \(\overrightarrow { a } -\overrightarrow { d } \) is parallel to \(\overrightarrow { b } -\overrightarrow { c } \) provided \(\overrightarrow { a } \neq \overrightarrow { d } \ and\ \overrightarrow { b } \neq \overrightarrow { c } \)

If \(\overrightarrow { a } =\hat { i } -\hat { j } +7\hat { k } \) and \(\overrightarrow { b } =5\hat { i } -\hat { j } +\lambda \hat { k } \) then find the value of \(\lambda\) so that the vectors \(\overrightarrow { a } +\overrightarrow { b } \ and\ \overrightarrow { a } -\overrightarrow { b } \) are orthogonal.

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Answers Key:

Generic 3

5 x 3 = 15

Clearly, \(\vec{b}+\vec{c}=\left(b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\right)+\left(c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}\right)\)
\(=\left(b_{1}+c_{1}\right) \hat{i}+\left(b_{2}+c_{2}\right) \hat{j}+\left(b_{3}+c_{3}\right) \hat{k}\)
\(\therefore \vec{a} \times(\vec{b}+\vec{c})=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1}+c_{1} & b_{2}+c_{2} & b_{3}+c_{3} \end{array}\right|\)
\(=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{array}\right|+\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right|\)
[by property of determinant]
\(=(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\)
Hence, \(\vec{a} \times(\vec{b}+\vec{c})=(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\)

Given,\(|\vec{r}|=14\) and if \(\vec{r}=a \hat{i}+b \hat{j}+c \hat{k}\) then \(a=2 \lambda, b=3 \lambda\) \(\text { and } c=-6 \lambda \text { for some } \lambda \neq 0 \text { . }\)
\(\therefore \text { Direction cosines } l, m \text { and } n \text { are } l=\frac{a}{|\vec{r}|}=\frac{2 \lambda}{14}=\frac{\lambda}{7}\)
\(m=\frac{b}{|\vec{r}|}=\frac{3 \lambda}{14} \text { and } n=\frac{c}{|\vec{r}|}=\frac{-6 \lambda}{14}=\frac{-3 \lambda}{7}\)
Also, we know that \(l^{2}+m^{2}+n^{2}=1\)
\( \therefore \frac{\lambda^{2}}{49}+\frac{9 \lambda^{2}}{196}+\frac{9 \lambda^{2}}{49}=1 \)
\(\Rightarrow \frac{4 \lambda^{2}+9 \lambda^{2}+36 \lambda^{2}}{196}=1 \)
\(\Rightarrow 49 \lambda^{2}=196 \Rightarrow \lambda^{2}=\frac{196}{49} \)
\(\lambda^{2}=4 \Rightarrow \lambda=\pm 2 \)
So, the direction cosines l,m and n are \(\frac{2}{7}, \frac{3}{7} \text { and } \frac{-6}{7}\) .
[\(\because \vec{r} \)makes an acute angle with X -axis, so we will take positive value of \(\lambda\)]
\(\because \ \vec{r}=\hat{r} \cdot|\vec{r}|\)
\(\therefore \vec{r}=(l \hat{i}+m \hat{j}+n \hat{k}) \cdot|\vec{r}|=\left(\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}-\frac{6}{7} \hat{k}\right) \cdot 14\)
\(=4 \hat{i}+6 \hat{j}-12 \hat{k}\)
Thus, the components of \(\vec{r} \text { are } 4 \hat{i}, 6 \hat{j} \text { and }-12 \hat{k}\).

\(\overset { \rightarrow }{ x } .\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) +\overset { \rightarrow }{ y } .\left( \overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) +\overset { \rightarrow }{ z } .\left( \overset { \rightarrow }{ c } +\overset { \rightarrow }{ a } \right) \)
\(\\ =\frac { \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } .\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) +\frac { \overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } .\left( \overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) +\frac { \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } .\left( \overset { \rightarrow }{ c } +\overset { \rightarrow }{ a } \right) \)
\(=\frac { 1 }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } \left[ (\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } ).\overset { \rightarrow }{ a } +(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } ).\overset { \rightarrow }{ b } +(\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } ).\overset { \rightarrow }{ b } +(\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } ).\overset { \rightarrow }{ c } +(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } ).\overset { \rightarrow }{ c } +(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } ).\overset { \rightarrow }{ a } \right] \)
\(=\frac { 1 }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } \left[ \left[ \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \overset { \rightarrow }{ a } \right] +\left[ \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \overset { \rightarrow }{ b } \right] +\left[ \overset { \rightarrow }{ c } \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \right] +\left[ \overset { \rightarrow }{ c } \overset { \rightarrow }{ a } \overset { \rightarrow }{ c } \right] +\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] +\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ a } \right] \right] \)
\(=\frac { 1 }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } \left[ \left[ \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \overset { \rightarrow }{ a } \right] +0+\left[ \overset { \rightarrow }{ c } \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \right] +0+\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] +0 \right] \)
\(=\frac { 3\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] }{ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] } =3.\)

\((\overrightarrow { a } -\overrightarrow { d } )\) x \((\overrightarrow { b } -\overrightarrow { c } )\)
\(=\overrightarrow { a } *\overrightarrow { b } -\overrightarrow { a } *\overrightarrow { c } -\overrightarrow { d } *\overrightarrow { b } +\overrightarrow { d } *\overrightarrow { c } \)
\(=\overrightarrow { c } *\overrightarrow { d } -\overrightarrow { b } *\overrightarrow { d } -\overrightarrow { d } *\overrightarrow { b } +\overrightarrow { d } -\overrightarrow { c } \)
\([\because \overrightarrow { a } *\overrightarrow { b } =\overrightarrow { c } -\overrightarrow { d } and\overrightarrow { a } *\overrightarrow { c } =\overrightarrow { b } *\overrightarrow { d } ]\)
\(=\overrightarrow { c } *\overrightarrow { d } -\overrightarrow { b } *\overrightarrow { d } +\overrightarrow { b } *\overrightarrow { d } -\overrightarrow { c } *\overrightarrow { d } =\overrightarrow { 0 } \)
Hence, \((\overrightarrow { a } -\overrightarrow { d } )\) is parallel to \((\overrightarrow { b } -\overrightarrow { c } )\)

Given, \(\vec{a}=\hat{i}-\hat{j}+7 \hat{k} \text { and } \vec{b}=5 \hat{i}-\hat{j}+\lambda \hat{k}\)
Now, \(\vec{a}+\vec{b}=6 \hat{i}-2 \hat{j}+(7+\lambda) \hat{k}\)
and \(\vec{a}-\vec{b}=-4 \hat{i}+(7-\lambda) \hat{k}\)
\(\because(\vec{a}+\vec{b}) \text { and }(\vec{a}-\vec{b})\) are orthogonal.
\(\begin{aligned} \therefore & (\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b}) =0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & {[6 \hat{i}-2 \hat{j}+(7+\lambda) \hat{k}] \cdot[-4 \hat{i}+(7-\lambda) \hat{k}] } =0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & -24+49-\lambda^2 =0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & \lambda^2 =25 \end{aligned}\)
\(\begin{aligned} \Rightarrow & \lambda = \pm 5 \end{aligned}\)