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Questions:

Generic 2

15 x 2 = 30

Prove that \(|\left[\begin{array}{lll} \vec{a} & \vec{b} & \bar{c} \end{array}\right]|=a b c\) if and only if \(\vec a,\vec{b}, \vec{c}\) are mutually perpendicular.

If \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0},|\vec{a}|=3,|\vec{b}|=5 \text { and }|\vec{c}|=7,\) find the angle between \(\vec a \text { and } \vec{b}\)

If \(\vec{a} \text { and } \vec{b}\), are unit vectors inclined at an angle \(\theta\), then prove that \(\sin \frac{\theta}{2}=\frac{1}{2}|\vec{a}-\vec{b}|\)

Find the projection of the vector \(7 \hat{i}+\hat{j}-4 \hat{k}\) on \(2 \hat{i}+6 \hat{j}+3 \hat{k}\)

For any vector \(\vec{r},\) prove that \(\vec{r}=(\vec{r} \cdot \hat{i}) \hat{i}+(\vec{r} \cdot \hat{j}) \hat{j}+(\vec{r} \cdot \hat{k}) \hat{k}\)

lf \(\vec a\)and \(\vec b\)are two vectors iuch that \(|\vec{a}|=4\), \(|\vec{b}|=3 \text { and } \vec{a} \cdot \vec{b}=6\). Find the angle between \(\vec a \text { and } \vec{b}\)

For what value of m the vectors \(\vec a\) and \(\vec b\) perpendicular to each other.
\((i) \ \vec{a}=m \hat{i}+2 \hat{j}+\hat{k}\ and \ \vec{b}=4 \hat{i}-9 \hat{j}+2 \hat{k} \)
\((ii) \ \vec{a}=5 \hat{i}-9 \hat{j}+2 \hat{k}\ and \ \vec{b}=m \hat{i}+2 \hat{j}+\hat{k}\)

Find \(\vec{a} \cdot \vec{b}\) when
\((i)\ \vec{a}=\hat{i}-2 \hat{j}+\hat{k}\ and\ \vec{b}=4 \hat{i}-4 \hat{j}+7 \hat{k} \)
\((ii)\ \vec{a}=\hat{j}+2 \hat{k}\ and\ \vec{b}=2 \hat{i}+\hat{k} \)
\((iii)\ \vec{a}=\hat{j}-2 \hat{k}\ and\ \vec{b}=2 \hat{i}+3 \hat{j}-2 \hat{k} \)

Prove that for any two vectors \(\vec{a} \text { and } \vec{b}\)\(|\vec{a}+\vec{b}| \leq|\vec{a}|+|b|\) (Triangle inequality)

10.

Let \(\overset { \rightarrow }{ a } ,\overset { \rightarrow }{ b } ,\overset { \rightarrow }{ c } \) be unit vectors such \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =\overset { \rightarrow }{ a } .\overset { \rightarrow }{ c } =0\) and the angle between \(\overset { \rightarrow }{ b } \) and \(\overset { \rightarrow }{ c } \) is \(\frac { \pi }{ 6 } \). Prove that \(\overset { \rightarrow }{ a } =\pm 2\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \)

11.

Find the equation of the plane containing the line of intersection of the planes x + y + Z - 6 = 0 and 2x + 3y + 4z + 5 = 0 and passing through the point (1, 1, 1)

12.

If the planes \({ \overset { \rightarrow }{ r } }.\left( \overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) =7\) and \({ \overset { \rightarrow }{ r } }.\left( \lambda \overset { \wedge }{ i } +2\overset { \wedge }{ j } -7\overset { \wedge }{ k } \right) =26\) are perpendicular. Find the value of λ.

13.

Find the parametric form of vector equation of the plane passing through the point (1, -1, 2) having 2, 3, 3 as direction ratios of normal to the plane.

14.

Find the parametric form of vector equation of a line passing through a point (2, -1, 3) and parallel to line \({ \overset { \rightarrow }{ r } }=\left( \overset { \wedge }{ i } +\overset { \wedge }{ j } \right) +t\left( 2\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \right) \)

15.

Find the Cartesian equation of a line passing through the points A(2, -1, 3) and B(4, 2, 1)

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Answers Key:

Generic 2

15 x 2 = 30

\(\vec a,\vec{b}, \vec{c}\)are mutually perpendicular \(\Leftrightarrow\left|\left[\begin{array}{lll} \vec{a}, \vec{b}, \vec{c} \end{array}\right]\right|\)
\( \Leftrightarrow|[\vec{a}, \vec{b}, \ \vec{c}]|=|\vec{a}||\vec{b}||\vec{c}| \)
\( \Leftrightarrow|[\vec{a}, \vec{b}, \ \vec{c}]|=a b c \)

\( \vec{a}+\vec{b}+\vec{c} =\overrightarrow{0} \)
\(\vec{a}+\vec{b} =-\vec{c} \)
\( (\vec{a}+\vec{b})^{2}=(-\vec{c})^{2} \)
\( \Rightarrow(\vec{a})^{2}+(\vec{b})^{2}+2 \vec{a} \cdot \vec{b}=(\vec{c})^{2} \)
\( \Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+\left.2|\vec{a}| \vec{b}|\cos \theta=| \vec{c}\right|^{2} \)
\( \Rightarrow 3^{2}+5^{2}+2(3)(5) \cos \theta=7^{2} \)
\( \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}\)

\( |\vec{a}-\vec{b}|^{2} =\vec{a}^{2}+\vec{b}^{2}-2 \vec{a} \cdot \vec{b}=1+1-2|\vec{a}||\vec{b}| \cos \theta \)
\( =2-2 \cos \theta=2(1-\cos \theta)=2\left(2 \sin ^{2} \frac{\theta}{2}\right) \)
\(\therefore|\vec{a}-\vec{b}| =2 \sin \frac{\theta}{2} \Rightarrow \sin \frac{\theta}{2}=\frac{1}{2}|\vec{a}-\vec{b}| \)

Let \(\vec{a}=7 \hat{i}+\hat{j}-4 \hat{k} ; \vec{b}=2 \hat{i}+6 \hat{j}+3 \hat{k}\)
Projection of \(\vec{a} \text { on } \vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\)
\( =\frac{(7 \hat{i}+\hat{j}-4 \hat{k}) \cdot(2 \hat{i}+6 \hat{j}+3 \hat{k})}{|2 \hat{i}+6 \hat{j}+3 \hat{k}|} \)
\( =\frac{14+6-12}{\sqrt{4+36+9}}=\frac{8}{7} \)

Let \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\) be an arbitrary vector
\( \vec{r} \cdot \hat{i}=(x \hat{i}+y \hat{j}+z \hat{k}) \cdot \hat{i}=x \)
\( \vec{r} \cdot \hat{j}=(x \hat{i}+y \hat{j}+z \hat{k}) \cdot \hat{j}=y \)
\( \vec{r} \cdot \hat{k}=(x \hat{i}+y \hat{j}+z \hat{k}) \cdot \hat{k}=z \)
\( (\vec{r} \cdot \hat{i}) \hat{i}+(\vec{r} \cdot \hat{j}) \hat{j}+(\vec{r} \cdot \hat{k}) \hat{k}=x \hat{i}+y \hat{j}+z \hat{k}=\vec{r}\)

\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{6}{(4)(3)}=\frac{1}{2} \ \)
\(\Rightarrow \theta=\frac{\pi}{3}\)

(i) Given \( \vec{a} \perp \vec{b} \)
\( \therefore \vec{a} \cdot \vec{b}=0 \Rightarrow(m \hat{i}+2 \hat{j}+\hat{k}) \cdot(4 \hat{i}-9 \hat{j}+2 \hat{k})=0 \)
\( \Rightarrow 4 m-18+2=0 \Rightarrow m=4 \)
(ii) \( (5 \hat{i}-9 \hat{j}+2 \hat{k}) \cdot(m \hat{i}+2 \hat{j}+\hat{k})=0 \)
\( \Rightarrow 5 m-18+2=0 \Rightarrow m=\frac{16}{5}\)

\((i) \ \vec{a} \cdot \vec{b}=(\hat{i}-2 \hat{j}+\hat{k}) \cdot(4 \hat{i}-4 \hat{j}+7 \hat{k}) \)
\(=(1)(4)+(-2)(-4)+(1)(7)=19 \)
\((ii) \ \vec{a} \cdot \vec{b}=(\hat{j}+2 \hat{k}) \cdot(2 \hat{i}+\hat{k}) \)
\(=(0)(2)+(1)(0)+(2)(1)=2 \)
\((iii) \ \vec{a} \cdot \vec{b}=(\hat{j}-2 \hat{k}) \cdot(2 \hat{i}+3 \hat{j}-2 \hat{k}) \)
\(=(0)(2)+(1)(3)+(-2)(-2)=7\)

We have
\( |\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+2(\vec{a} \cdot \vec{b}) \)
\(\Rightarrow|\vec{a}+\vec{b}|^{2} =|\vec{a}|^{2}+|\vec{b}|^{2}+2|\vec{a}||\vec{b}| \cos \theta \)
\( \leq|\vec{a}|^{2}+|\vec{b}|^{2}+2|\vec{a} \| \vec{b}|\ \ [\because \cos \theta \leq 1]\)
\( \Rightarrow|\vec{a}+\vec{b}|^{2} \leq(|\vec{a}|+|\vec{b}|)^{2} \)
\( \Rightarrow|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|\)

10.

Given \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =\overset { \rightarrow }{ a } .\overset { \rightarrow }{ c } =0\)] ⇒ \(\overset { \rightarrow }{ a } \) 丄 \(\overset { \rightarrow }{ b } \)and \(\overset { \rightarrow }{ a } \)丄 \(\overset { \rightarrow }{ c } \)
⇒ \(\overset { \rightarrow }{ a } \) 丄^r to the plane containing \(\overset { \rightarrow }{ b } \)and \(\overset { \rightarrow }{ c } \)
Also, \(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } =\left| \overset { \rightarrow }{ b } \right| \left| \overset { \rightarrow }{ c } \right| \) Since \(\overset { \wedge }{ n } \) [where θ is the angle between \(\overset { \rightarrow }{ b } \)and \(\overset { \rightarrow }{ c } \)]
= 1 \(\times\) 1. sin \(\frac { \pi }{ 6 } \).\(\overset { \rightarrow }{ a } \)
[Since \(\overset { \rightarrow }{ b } \) and \(\overset { \rightarrow }{ c } \) are unit vectors \(\overset { \rightarrow }{ a } \) 丄 both \(\overset { \rightarrow }{ b } \)and \(\overset { \rightarrow }{ c } \) \(\overset { \rightarrow }{ { n } } \) = \(\overset { \rightarrow }{ a } \)]
\(=\frac { 1 }{ 2 } \overset { \rightarrow }{ a } \)
\( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } =\frac { 1 }{ 2 } \overset { \rightarrow }{ a } \)
\(\Rightarrow \overset { \rightarrow }{ a } =\pm 2\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \)

11.

The equation of the required plane through the intersection of the given planes is
( x + y + z - 6 ) + λ (2x + 3y + 4z + 5) = 0 ......(1)
This passes through (1, 1, 1)
∴ ( 1+ 1 + z - 6) + λ (2 + 3+ 4 + 5) = 0
⇒ -3 +14λ = 0 \(\Rightarrow \lambda =\frac { 3 }{ 14 } \)
Substituting \(\lambda =\frac { 3 }{ 14 } \) in (1) we get
( x + y + z - 6 )+\(\frac { 3 }{ 14 } \) (2x + 3y + 4z + 5) = 0
⇒ 14( x + y + z - 6 ) +3 (2 + 3+ 4 + 5) = 0
⇒ 20x + 23y + 26z - 69 = 0

12.

The planes \(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ { n }_{ 1 } } ={ d }_{ 1 }\) and \(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ { n }_{ 2 } } ={ d }_{ 2 }\) are perpendicular if \(\overset { \rightarrow }{ { n }_{ 1 } } .\overset { \rightarrow }{ { n }_{ 2 } } =0\)
Here \(\overset { \rightarrow }{ { n }_{ 1 } } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ { n }_{ 2 } } =\lambda \overset { \wedge }{ i } +2\overset { \wedge }{ j } -7\overset { \wedge }{ k } \)
\(\therefore \overset { \rightarrow }{ { n }_{ 1 } } .\overset { \rightarrow }{ { n }_{ 2 } } =\left( \overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) .\left( \lambda \overset { \wedge }{ i } +2\overset { \wedge }{ j } -7\overset { \wedge }{ k } \right) =0\)
⇒ λ + 4 - 21 = 0
⇒ λ - 17 = 0
⇒ λ = 17

13.

Since the plane passing through the point \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \) and is normal to the vector \(\overset { \rightarrow }{ n } =2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \)
the vector equation of the plane is \(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ n } =\overset { \rightarrow }{ a } .\overset { \rightarrow }{ n } \)
\(\Rightarrow { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) =\left( \overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) .\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)
\(\Rightarrow { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)= 2 - 3 + 4 = 3
\(\therefore { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)= 3

14.

The parametric form of vector equation of a line passing through a point \(\left( \overset { \rightarrow }{ a } \right) \) and parallel to \(\overset { \rightarrow }{ b } \) is
\({ \overset { \rightarrow }{ r } }=\overset { \rightarrow }{ a } +t\overset { \rightarrow }{ b } \), t ∈ R
⇒ Here \(\overset { \rightarrow }{ a } =2\overset { \wedge }{ i } -\overset { \wedge }{ j } +3\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ b } =2\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \)
\(\therefore { \overset { \rightarrow }{ r } }=\left( 2\overset { \wedge }{ i } -\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) +t\left( 2\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \right) \), t ∈ R which is the required equation of a line.

15.

Given (x₁, y₁, z₁) is (2, -1, 3) (x₂, y₂, z₂) is (4, 2, 1)
Cartesian equation of a line passing through two points is \(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \)
\(\Rightarrow \frac { x-2 }{ 2 } =\frac { y+1 }{ 3 } =\frac { z-3 }{ -2 } \)