Vector Algebra

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} \begin{array}{l} \{(\mathbf{a}+\mathbf{b}) \times(\mathbf{a}+\mathbf{c})\} \times(\mathbf{b} \times \mathbf{c}) \cdot(\mathbf{b}+\mathbf{c}) \\ \\ =(\mathbf{a} \times \mathbf{c}+\mathbf{b} \times \mathbf{a}+\mathbf{b} \times \mathbf{c}) \times(\mathbf{b} \times \mathbf{c}) \cdot(\mathbf{b}+\mathbf{c}) \\ \begin{array}{r} \{(\mathbf{a} \times \mathbf{c}) \times(\mathbf{b} \times \mathbf{c})+(\mathbf{b} \times \mathbf{a}) \times(\mathbf{b}+\mathbf{c})+(\mathbf{b} \times \mathbf{c}) \\ \times(\mathbf{b} \times \mathbf{c})\} \cdot(\mathbf{b}+\mathbf{c}) \end{array} \\ =\{[\mathbf{a} \mathbf{c} \mathbf{c}] \mathbf{b}-[\mathbf{a} \mathbf{c} \mathbf{b}] \mathbf{c}+[\mathbf{b} \mathbf{a} \mathbf{c}] \mathbf{b}-[\mathbf{b} \mathbf{a} \mathbf{b}] \mathbf{c}\} \cdot(\mathbf{b}+\mathbf{c}) \end{array} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\{-[\mathbf{a} \mathbf{c} \mathbf{b}] \mathbf{c}+[\mathbf{b} \mathbf{a} \mathbf{c}] \mathbf{b}\} \cdot(\mathbf{b}+\mathbf{c}) \\ & =-[\mathbf{a} \mathbf{c} \mathbf{b}]\{\mathbf{c} \cdot(\mathbf{b}+\mathbf{c})\}+[\mathbf{b} \mathbf{a} \mathbf{c}]\{\mathbf{b} \cdot(\mathbf{b}+\mathbf{c})\} \\ & =[\mathbf{a} \mathbf{b} \mathbf{c}]\left(\mathbf{c} \cdot \mathbf{b}+|\mathbf{c}|^2\right)-[\mathbf{a} \mathbf{b} \mathbf{c}]\left(|\mathbf{b}|^2+\mathbf{b} \cdot \mathbf{c}\right) \\ & =[\mathbf{a} \mathbf{b} \mathbf{c}]\left(|\mathbf{c}|^2-|\mathbf{b}|^2\right) \\ & =[\mathbf{a} \mathbf{b} \mathbf{c}] \times 0=0 \quad[\because|\mathbf{b}|=|\mathbf{c}|] \end{aligned} $

To compute the scalar triple product $[\mathbf{a} \mathbf{b} \mathbf{c}]$, we first use the determinant representation: $[a b c] = \left| \begin{array}{ccc} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{array} \right|$.

This can be expanded as follows:

$ [a b c] = 1 \left( (1 \cdot (1 + x - y)) - (x \cdot (1 - x)) \right) - 1 \left( x^2 - y \right) $

Breaking it down further:

$ = 1 \cdot (1 + x - y - x + x^2) - (x^2 - y) $

Simplifying the expression:

$ = (1 + x - y - x + x^2) - x^2 + y $

$ = 1 $

The result $1$ is independent of the variables $x$ and $y$. Therefore, the value of the scalar triple product $[\mathbf{a} \mathbf{b} \mathbf{c}]$ does not depend on $x$ or $y$.

The projection of a vector $\mathbf{a}$ on another vector $\mathbf{b}$ is calculated using the formula:

$ |\mathbf{a}| \cos \theta = \frac{(\mathbf{a} \cdot \mathbf{b})}{|\mathbf{b}|} $

Consider the line joining the points $ (3, 4, 5) $ and $ (4, 6, 3) $ as $ L_1 $, and the line joining the points $ (-1, 2, 4) $ and $ (1, 0, 5) $ as $ L_2 $.

The direction vector $ L_1 $ is derived as follows:

$ L_1 = (4 - 3)\hat{\mathbf{i}} + (6 - 4)\hat{\mathbf{j}} + (3 - 5)\hat{\mathbf{k}} = \hat{\mathbf{i}} + 2\hat{\mathbf{j}} - 2\hat{\mathbf{k}} $

Similarly, the direction vector $ L_2 $ is:

$ L_2 = (1 - (-1))\hat{\mathbf{i}} + (0 - 2)\hat{\mathbf{j}} + (5 - 4)\hat{\mathbf{k}} = 2\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + \hat{\mathbf{k}} $

The projection of $ L_1 $ onto $ L_2 $ is calculated by:

$ \text{Projection of } L_1 \text{ on } L_2 = \frac{L_1 \cdot L_2}{|L_2|} $

Calculating the dot product $ L_1 \cdot L_2 $:

$ L_1 \cdot L_2 = (1 \times 2) + (2 \times -2) + (-2 \times 1) = 2 - 4 - 2 = -4 $

The magnitude of $ L_2 $ is:

$ |L_2| = \sqrt{(2)^2 + (-2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 $

Thus, the projection is:

$ \frac{-4}{3} $

Therefore, the magnitude of the projection is:

$ \frac{4}{3} $

Let $c=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}}$

Given that $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are coplanar

$\begin{aligned} & \therefore [\mathbf{a} \mathbf{b} \mathbf{c}]=0 \\ & \Rightarrow \quad \left|\begin{array}{ccc} 2 & 1 & 1 \\ 1 & 2 & -1 \\ x & y & z \end{array}\right|=0 \\ & \Rightarrow x(-1-2)-y(-2-1)+z(4-1)=0 \end{aligned}$

$\Rightarrow \quad x-y-z=0$ .... (i)

$\because \mathrm{c}$ is perpendicular to a.

$\because \quad \mathbf{a} \cdot \mathbf{c}=0 \Rightarrow 2 x+y+z=0$ .... (ii)

From Eqs. (i) and (ii), we get

$\begin{aligned} & \frac{x}{0}=\frac{y}{3}=\frac{z}{-3}=\lambda \\ & \Rightarrow x=0, y=3 \lambda \text { and } z=-3 \lambda \\ & \text { But, }|c|=1 \Rightarrow x^2+y^2+z^2=1 \\ & \Rightarrow \quad 9 \lambda^2+9 \lambda^2=1 \Rightarrow \lambda= \pm \frac{1}{3 \sqrt{2}} \\ & \therefore \quad c= \pm \frac{1}{\sqrt{2}}(-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \end{aligned}$

Thus, we have $\mathbf{c}=\frac{1}{\sqrt{2}}(-\hat{\mathbf{j}}+\hat{k})$

Given that, $\left| {{{\widehat u + \widehat v} \over 2} + \widehat u \times \widehat v} \right| = 1$

$ \Rightarrow {\left| {{{\widehat u + \widehat v} \over 2} + \widehat u \times \widehat v} \right|^2} = 1$

$ \Rightarrow {{2 + 2\cos \theta } \over 4} + {\sin ^2}\theta = 1$ [$\because$ $\widehat u\,.\,(\widehat u \times \widehat v) = \widehat v(\widehat u \times \widehat v) = 0$]

$ \Rightarrow {\cos ^2}{\theta \over 2} = \cos 2\theta $

$ \Rightarrow \theta = n\pi \, \pm \,{\theta \over 2},\,n \in z$

$ \Rightarrow \theta = {{2\pi } \over 3}$

$ \Rightarrow \left| {\widehat u \times \widehat v} \right| = \sin {{2\pi } \over 3} = \sin {\pi \over 3} = \left| {{{\widehat u - \widehat v} \over 2}} \right|$

To determine if the points with position vectors $10\widehat{i} + 3\widehat{j}$, $12\widehat{i} - 5\widehat{j}$, and $a\widehat{i} + 11\widehat{j}$ are collinear, we need to check if the vectors between these points are scalar multiples of each other. This is equivalent to checking if the determinant of the matrix formed by these points is zero.

Let's consider the vectors formed by the points:

• Vector 1, $\mathbf{AB}$: from $10\widehat{i} + 3\widehat{j}$ to $12\widehat{i} - 5\widehat{j}$
• Vector 2, $\mathbf{AC}$: from $10\widehat{i} + 3\widehat{j}$ to $a\widehat{i} + 11\widehat{j}$

We can find these vectors as follows:

$\mathbf{AB} = (12 - 10)\widehat{i} + (-5 - 3)\widehat{j}$

$\mathbf{AB} = 2\widehat{i} - 8\widehat{j}$

$\mathbf{AC} = (a - 10)\widehat{i} + (11 - 3)\widehat{j}$

$\mathbf{AC} = (a - 10)\widehat{i} + 8\widehat{j}$

For the points to be collinear, $\mathbf{AB}$ and $\mathbf{AC}$ must be linearly dependent, such that:

$\mathbf{AB} = k \cdot \mathbf{AC}$ for some scalar $k$

Equating the components:

$2\widehat{i} - 8\widehat{j} = k \cdot \left((a - 10)\widehat{i} + 8\widehat{j}\right)$

Separating the components gives us two equations:

1. $2 = k(a - 10)$

2. $-8 = k \cdot 8$

From the second equation:

$k = -1$

Substituting $k = -1$ into the first equation:

$2 = -1 \cdot (a - 10)$

$2 = -a + 10$

Solving for $a$:

$a = 10 - 2$

$a = 8$

Thus, the points will be collinear if $a = 8$.

We have, | a | = 3, | b | = 4 and | c | = 5.

It is given that

a $ \bot $ (b + c), b $ \bot $ (c + a) and c $ \bot $ (a + b)

$\Rightarrow$ a . (b + c) = 0, b . (c + a) = 0 and c . (a + b) = 0

$\Rightarrow$ a . b + a . c = b . c + b . a = c . a = c . b = 0

or, a . b + b . c + c . a = 0

(adding all the above equations)

Now, | a + b + c |² = | a |² + | b |² + | c |² + 2 (a . b + b . c + c . a)

= 3² + 4² + 5² = 50

$\therefore$ | a + b + c | = 5$\sqrt2$

We have, $\left| {(a \times b).\,c} \right| = \left| a \right|\left| b \right|\left| c \right|$

$ \Rightarrow \left\| a \right\|\left\| b \right\|\left| c \right|\left. {\sin \theta \cos \alpha } \right| = \left| a \right|\left| b \right|\left| c \right|$

$ \Rightarrow \left| {\sin \theta } \right|\left| {\cos \alpha } \right| = 1 \Rightarrow \theta = {\pi \over 2}$ and $\alpha = 0$

$ \Rightarrow a \bot b$ and $c||\widehat n \Rightarrow a \bot b$ and c $\bot$ both a and b.

$\Rightarrow$ a, b, c are mutually perpendicular.

$\Rightarrow$ a . b = b . c = c . a = 0

We have given,

| a | = 1, | b | = 4, a . b = 2

Now, c = (2a $\times$ b) $-$ 3b

Multiplying by b both sides, we get

$b\,.\,c = (2a \times b)\,.\,b - 3|b{|^2}$

$ \Rightarrow b\,.\,c = 0 - 3{(4)^2} = - 48$

and $|c{|^2}\, = \,|(2a \times b)\,.\,3b{|^2}$

$|c{|^2}\, = \,{(2a \times b)^2} + 9|b{|^2} = 4{(a \times b)^2} + 9{(4)^2}$

$ \Rightarrow |c{|^2}\, = 4(|a{|^2}\,.\,|b{|^2} - {(a\,.\,b)^2}) + 144$

$ = 4(16 - 4) + 144 = 48 + 144 = 192$

$\therefore$ $b\,.\,c = |b||c|cos\theta $

$ \Rightarrow cos\theta = {{b\,.\,c} \over {|b||c|}} = {{ - 48} \over {4 \times \sqrt {192} }} = - {{\sqrt 3 } \over 2}$

$\theta = \pi - {\pi \over 6} = {{5\pi } \over 6}$

We have,

Vector r is coplanar with vector a and b

$\therefore$ $r = \lambda (b \times (a \times b))$

$ \Rightarrow r = \lambda ((b\,.\,b)a - (b\,.\,a)b)$

$ \Rightarrow r = \lambda (|b{|^2}a - (a\,.\,b)b)$

$ \Rightarrow r = \lambda [5a + b]$ [$\because$ $|b| = \sqrt {{2^2} + {1^2}} = \sqrt 5 $]

and $a\,.\,b = ( - \widehat i + \widehat j + \widehat k)\,.\,(2\widehat i + \widehat k) = - 1$

Given, $r\,.\,a = 7$

$\therefore$ $\lambda (5a + b)\,.\,a = 7$

$ \Rightarrow \lambda [5|a{|^2} + a\,.\,b] = 7$

$ \Rightarrow \lambda [15 - 1] = 7$ [$\because$ $|a{|^2} = 1 + 1 + 1 = 3$]

$ \Rightarrow \lambda = {1 \over 2}$

$\therefore$ $r = {1 \over 2}(5a + b)$

$r = {1 \over 2}( - 5\widehat i + 5\widehat j + 5\widehat k + 2\widehat i + \widehat k)$

$r = {{ - 3} \over 2}\widehat i + {5 \over 2}\widehat j + {6 \over 2}\widehat k$

z component of $r = {6 \over 2} = 3$