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Published on: 29/08/2019
Vector Algebra
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1.
Write the value of p for which \(\overrightarrow { a } =3\hat { i } +2\hat { j } +9\hat { k } \quad and\quad \overrightarrow { b } =\hat { i } +p\hat { j } +3\hat { k } \) are parallel vectors.
2.
Find the projection of \(\overrightarrow { a } \ on\ \overrightarrow { b } \) if \(\overrightarrow { a } .\overrightarrow { b } =8\ and\ \overrightarrow { b } =2\overrightarrow { i } +6\overrightarrow { j } +3\overrightarrow { k } \)
3.
Find a vector in the direction of \(\overrightarrow { a } =\overrightarrow { i } -2\overrightarrow { j } \) whose magnitude is 7.
4.
If \(\left| \overrightarrow { a } \right| =\sqrt { 3 } ,\left| \overrightarrow { b } \right| =2\) and angle between \(\overrightarrow { a } \ and\ \overrightarrow { b } \) is 60o find \(\overrightarrow { a } .\overrightarrow { b } \)
5.
If \(\overrightarrow { a } =\hat { i } +2\hat { j } -3\hat { k } \) and \(\overrightarrow { b } =2\hat { i } +4\hat { j } +9\hat { k } \) find a unit vector parallel to \(\overrightarrow { a } +\overrightarrow { b } \)
6.
If \(\left| \overrightarrow { a } \right| =2,\left| \overrightarrow { b } \right| =\sqrt { 3 } \) and \(\overrightarrow { a } .\overrightarrow { b } =\sqrt { 3 } \) find the angle between \(\overrightarrow { a } \ and\ \overrightarrow { b } \)
7.
If \(\overrightarrow { a } =\overrightarrow { i } +2\overrightarrow { j } -\overrightarrow { k } \ and\ \overrightarrow { b } =3\overrightarrow { i } +\overrightarrow { j } -5\overrightarrow { k } \) find a unit vector in the direction of \(\overrightarrow { a } -\overrightarrow { b } \)
8.
For what value of \(\lambda\) are the vectors \(\overrightarrow { a } =2\overrightarrow { i } +\lambda \overrightarrow { j } +\overrightarrow { k } \) and \(\overrightarrow { b } =\overrightarrow { i } -2\overrightarrow { j } +3\overrightarrow { k } \) perpendicular to each other?
9.
Find a unit vector in the direction of \(\overrightarrow { a } =3\overrightarrow { i } -2\overrightarrow { j } +6\overrightarrow { k } \)
10.
For any two vectors \(\vec{a} \text { and } \vec{b}\)
\(\vec{a}+\vec{b}=\vec{b}+\vec{a}\) (Commutative property)
1.
\(\frac{3}{1}=\frac{2}{p}=\frac{9}{3} \Rightarrow p=\frac{2}{3}\)
2.
\(\text { Projection of } \vec{a} \text { on } \vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\frac{8}{7}\)
3.
Vector of magnitude 7 along is \(7 \hat{a}=\frac{7(\hat{i}-2 \hat{j})}{\sqrt{5}}\)
\(7\hat { a } =\frac { 7 }{ \sqrt { 5 } } \hat { i } -\frac { 14 }{ \sqrt { 5 } } \hat { j } \)
4.
\(\cos 60^{\circ}=\frac{\vec{a} \cdot \vec{b}}{2 \sqrt{3}} \Rightarrow \frac{1}{2}=\frac{\vec{a} \cdot \vec{b}}{2 \sqrt{3}} \Rightarrow \vec{a} \cdot \vec{b}=\sqrt{3}\)
5.
\(\frac { 1 }{ 3 } \hat { i } +\frac { 2 }{ 3 } \hat { j } +\frac { 2 }{ 3 } \hat { k } \)
6.
\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a} \| \vec{b}|}=\frac{\sqrt{3}}{2 \sqrt{3}}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}\)
7.
\(\vec{r}=\vec{a}-\vec{b}=-2 \hat{i}+\hat{j}+4 \hat{k},|\vec{r}|=\sqrt{4+1+16}=\sqrt{21} \)
\(\text { Now, unit vector } \hat{r}=\frac{\vec{r}}{|\vec{r}|}=\frac{-2}{\sqrt{21}} \hat{i}+\frac{1}{\sqrt{21}} \hat{j}+\frac{4}{\sqrt{21}} \hat{k} \text { ; } \)
8.
\(\text { If } a \text { and } \vec{b} \text { are perpendicular, then } \vec{a} \cdot \vec{b}=0\)
\(\Rightarrow 2-2 \lambda+3=0 \Rightarrow \lambda=\frac{5}{2}\)
9.
\(\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{3 \hat{i}-2 \hat{j}+6 \hat{k}}{\sqrt{9+4+36}}=\frac{3}{7} \vec{i}-\frac{2}{7} \vec{j}+\frac{6}{7} \vec{k}\)
10.
Consider the parallelogram ABCD. Let \(\overrightarrow{\mathrm{AB}}=\vec{a} \text { and } \overrightarrow{\mathrm{BC}}=\vec{b}\) then using the triangle law, from triangle ABC, we have \(\overrightarrow{\mathrm{AC}}=\vec{a}+\vec{b}\)
Now, since the opposite sides of a parallelogram are equal and parallel, from figure \(\overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{BC}}=\vec{b}\) \(\overrightarrow{\mathrm{DC}}=\overrightarrow{\mathrm{AB}}=\vec{a} .\)
Again using triangle law, from triangle ADC, we have
\( \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DC}}=\vec{b}+\vec{a} \\ \vec{a}+\vec{b}=\vec{b}+\vec{a} \)
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