Vectors from a geometric viewpoint - class-XI

The expression involves a complex vector product. Given the properties of cross products and dot products, the result of such a complex vector identity is generally not equal to 1 unless specific conditions are met. This is a false statement.

$\begin{array}{l} \overrightarrow { a } =2\left( { { { \log   } _{ 3 } }x,a } \right) ,\overrightarrow { b } =\left( { -3,a{ { \log   } _{ 3 } }x,{ { \log   } _{ 3 } }x } \right)  \ \overrightarrow { a } \overrightarrow { b } =\left| a \right| \left| b \right| \cos  \theta  \ \cos  \theta =\frac { { \overrightarrow { a } .\overrightarrow { b }  } }{ { \left| a \right| \left| b \right|  } }  \ =\frac { { -6+a{ { \left( { { { \log   } _{ 3 } }x } \right)  }^{ 2 } }+a.{ { \log   } _{ 3 } }x } }{ { \sqrt { 4+{ { \left( { { { \log   } _{ 3 } }x } \right)  }^{ 2 } }+{ a^{ 2 } } } .\sqrt { 9+{ a^{ 2 } }{ { \left( { { { \log   } _{ 3 } }x } \right)  }^{ 2 } }+{ { \left( { { { \log   } _{ 3 } }x } \right)  }^{ 2 } } }  } }  \ for, , acute, , angle, , a>0 \ Option, , , C, , is, , correct. \end{array}$

$\displaystyle a\times b=a\times c\Rightarrow a\times b-a\times c=0$
$\Rightarrow a\times (b-c)=0$ $\Rightarrow a \parallel (b-c)$
$\Rightarrow b-c = \lambda a\Rightarrow b= c+\lambda a$

$\displaystyle a\times \left ( b+c \right )+b\times \left ( c+a \right )+c\times \left ( a+b \right )$
$=a\times b+a\times c+b\times c+b\times a+c\times a+c\times b$
$=a\times b-c\times a+b\times c-a\times b+c\times a-b\times c = 0$

$\vec r \times \vec a = \vec b \times \vec a\Rightarrow \vec{r} = \vec{b}+\lambda \vec{a}$
and $\vec r \times \vec b = \vec a \times \vec b\Rightarrow \vec{r} = \vec{a}+\mu \vec{b}$
For intersection of both the lines, $\vec{b}+\lambda \vec{a}=\vec{a}+\mu \vec{b}$
Comparing coefficients, $\lambda=\mu = 1$
Hence point of intersection is $\vec{r}=\vec{a}+\vec{b} = 3\hat{i}+\hat{j}-\hat{k}$

We get 

Given,

$|\bar{a}|=1, |\bar{b}|=2, (\bar{a}, \bar{b})=\dfrac{2z}{3}$

$\vec{r} \times \vec{a} = \vec{b} \times \vec{a}$ or $(\vec{r} - \vec{b}) \times \vec{a} = 0$
$\vec{r} \times \vec{b} = \vec{a} \times \vec{b}$ or $(\vec{r} - \vec{a}) \times \vec{b} = 0$
If $\vec{r} = x \widehat{i} + y \widehat{j} + z\widehat{k}$, then 

$\bar{a}+p\bar{b}+q\bar{c}=0$
$\bar{b}=\dfrac{-q\bar{c}-\bar{a}}{p}$

By triangle law, $\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =\overrightarrow { 0 } $
Taking cross product by $\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } $ respectively 
$\overrightarrow { a } \times \left( \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c }  \right) =\overrightarrow { a } \times \overrightarrow { 0 } =\overrightarrow { 0 } $
$\Rightarrow \overrightarrow { a } \times \overrightarrow { a } \times \overrightarrow { a } \times \overrightarrow { b } +\overrightarrow { a } \times \overrightarrow { c } =\overrightarrow { a } $
$\Rightarrow \overrightarrow { a } \times \overrightarrow { b } =\overrightarrow { c } \times \overrightarrow { a } \quad \left[ \because \overrightarrow { a } \times \overrightarrow { a } =\overrightarrow { 0 }  \right] $
Similarly, $\overrightarrow { a } \times \overrightarrow { b } =\overrightarrow { b } \times \overrightarrow { c. } $
$\therefore \overrightarrow { a } \times \overrightarrow { b } =\overrightarrow { b } \times \overrightarrow { c } =\overrightarrow { c } \times \overrightarrow { a. } $

Given,  $\displaystyle a\cdot b=a\cdot c$ and $\displaystyle a\times b=a\times c,$
$\Rightarrow a\cdot (b-c) = 0$ and $a\times (b-c) = 0$
Hence  either $a=0$ or $b=c$

$\displaystyle \overrightarrow{a}=\hat{i}+\hat{j}$ & $\displaystyle \hat{b}=\hat{2i}-\hat{k}:\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{b}\times \overrightarrow{a}\Rightarrow \left ( \overrightarrow{r}-\overrightarrow{b} \right )\times \overrightarrow{a}=0
\Rightarrow \overrightarrow{r}=\overrightarrow{b}+\lambda \overrightarrow{a}$ 

Given, $\displaystyle \vec{R}\times \vec{B}=\vec{C}\times \vec{B} \Rightarrow \vec{R}\times \vec{B}-\vec{C}\times \vec{B}=0$
$\Rightarrow (\vec{R}-\vec{C})\times \vec{B}=0\Rightarrow R = \vec{C}+k\vec{B}$
Also given, $\vec{R}\cdot \vec{A} =0\Rightarrow k =- \cfrac{\vec{C}\cdot \vec{A}}{\vec{B}\cdot\vec{A}}=-\cfrac{15}{3}=-5$
Hence $\vec{R} =\vec{C}-5\vec{B}=-\hat{i}-8\hat{j}+2\hat{k} $ 

Given $\vec b = \widehat i + 2 \widehat j + \widehat k $ & $ \vec c = 3 \widehat i + 2 \widehat k $
$\vec r \times \vec b = \vec r \times \vec c$
$(\vec r \times \vec b) - (\vec r \times \vec c) = \vec 0 $
$\Rightarrow \vec r \times (\vec b - \vec c) = \vec 0$
$\Rightarrow \vec r = \lambda (\vec b - \vec c) = \lambda (- 2 \widehat i + 2 \widehat j - \widehat k)$
$\Rightarrow \displaystyle \widehat r = \pm \left ( \frac{2 \widehat i - 2 \widehat j + \widehat k}{3} \right )$

$\vec r \times \vec a = \vec b \times \vec a\Rightarrow \vec{r} = \vec{b}+\lambda \vec{a}$
and $\vec r \times \vec b = \vec a \times \vec b\Rightarrow \vec{r} = \vec{a}+\mu \vec{b}$
For intersection of both the lines, $\vec{b}+\lambda \vec{a}=\vec{a}+\mu \vec{b}$
Comparing coefficients, $\lambda=\mu = 1$
Hence point of intersection is $\vec{r}=\vec{a}+\vec{b} = 3\hat{i}+\hat{j}-\hat{k}$

$\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{b}\times\overrightarrow{c}$
$(\overrightarrow{a}+\overrightarrow{c})\times\overrightarrow{b}=0$
$\Rightarrow (\overrightarrow{a}+\overrightarrow{c})\parallel \overrightarrow{b}$

Given a cross b = 2a cross c, we can write a cross (b - 2c) = 0. This implies (b - 2c) is parallel to a, so b - 2c = lambda a. Using the magnitude squared of (b - 2c) = lambda^2 * |a|^2, we expand |b|^2 + 4|c|^2 - 4(b dot c) = lambda^2. With |b|=4, |c|=1, and cos(theta)=1/4, we get 16 + 4 - 4(411/4) = 20 - 4 = 16 = lambda^2, so lambda = +/- 4.

$\vec{a}\times \vec{b}= \vec{c}\times \vec{d}$
$\vec{a}\times \vec{c}= \vec{b}\times \vec{d}$
$\vec{a}\times (\vec{b}-\vec{c})= (\vec{c}-\vec{b})\times \vec{d}$
$(\vec{a}-\vec{d})\times (\vec{b}-\vec{c})= 0$
$(\vec{a}-\vec{d})\ \parallel \ (\vec{b}-\vec{c})$

$\vec{a}\times \vec{b}=\vec{b}\times \vec{c}$
$(\vec{a}+\vec{c})\times \vec{b}=o$
$(\vec{a+\vec{c}}+\vec{b})\times \vec{b}=\vec{b\times \vec{b}}$
$(\vec{a}+\vec{c}+\vec{b})\times \vec{b}=0$
$\vec{a}+\vec{c}+\vec{b}=\vec{0}$   hence $\vec{a},\vec{b},\vec{c}$ are non collinear

$\vec{a}\times \vec{b}=\vec{c}\times \vec{d}  -(1)$
$\vec{a}\times \vec{c}=\vec{b}\times \vec{d}  -(2)$
$(1) - (2)$
$\vec{a}\times \vec{b}+\vec{c}\times \vec{a}=\vec{c}\times \vec{d}+\vec{d}\times \vec{b}$
$(\vec{a}-\vec{d})\times \vec{b}=c\times (\vec{d}-\vec{a})$
$(\vec{a}-\vec{d})\times \vec{b}+(\vec{d}-\vec{a})\times \vec{c}=0$
$(\vec{a}-\vec{d})\times (\vec{b}-\vec{c})=0$
which mean $(\vec{a}-\vec{d})||(\vec{b}-\vec{c})$

Let $r,a,b$ and $c$ be vectors.
It is given that
$r\times a=b\times a$
$r\times a-(b\times a)=0$
$(r-b)\times a=0$
Hence $r-b$ is a vector parallel to vector $a$.
$r=b+\mu a$ ...(i)
It is given that $r.c=0$.
Hence $r$ vector is perpendicular to $c$ vector.
$(b+\mu a).c=0$ ...(from i)
$b.c+\mu(a.c)=0$
$\mu(a.c)=-b.c$
$\mu=-\dfrac{b.c}{a.c}$
Hence $r=b-\dfrac{b.c}{a.c}a$

Given, $\left| a \right| =\left| b \right| =1$ and $a\cdot b=\cos { \dfrac { \pi  }{ 3 }  } $
Consider,
$\left{ a\times \left( b+a\times b \right)  \right} \cdot b=\left{ a\times b+a\times \left( a\times b \right)  \right} \cdot b$
            $=\left( a\times b \right) \cdot b+\left{ \left( a\cdot b \right) \cdot a-\left( a\cdot a \right) \cdot b \right} \cdot b$
            $=\left[ \begin{matrix} a & b & b \end{matrix} \right] +{ \left( a\cdot b \right)  }^{ 2 }-{ \left| a \right|  }^{ 2 }{ \left| b \right|  }^{ 2 }$
            $=0+\cos ^{ 2 }{ \dfrac { \pi  }{ 3 }  } -1=\dfrac { 1 }{ 4 } -1=\dfrac { -3 }{ 4 } $

$\vec { \lambda  } +\vec { \mu  } =\vec { a } \times \left( \vec { b } +\vec { c }  \right) +\vec { b } \times \left( \vec { c } +\vec { a }  \right) $

$\vec{r}\times \vec{a}=\vec{b}\times \vec{a}$

$\left( \overrightarrow{c}+\overrightarrow{b} \right )\times \left ( \overrightarrow{c}+\overrightarrow{a} \right ).\left ( \overrightarrow{c}+\overrightarrow{b}+\overrightarrow{a} \right)$

$= \left(\overrightarrow{c} \times \overrightarrow{a}+\overrightarrow{b} \times \overrightarrow{c} + \overrightarrow{b}\times \overrightarrow{a}\right)\cdot \left ( \overrightarrow{c}+\overrightarrow{b}+\overrightarrow{a} \right)$
$= \overrightarrow{c} \times \overrightarrow{a}\cdot \overrightarrow{c}+\overrightarrow{b} \times \overrightarrow{c}\cdot \overrightarrow{c} + \overrightarrow{b}\times \overrightarrow{a}\cdot \overrightarrow{c}$
$+\overrightarrow{c} \times \overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b} \times \overrightarrow{c}\cdot \overrightarrow{b} + \overrightarrow{b}\times \overrightarrow{a}\cdot \overrightarrow{b}$
$+\overrightarrow{c} \times \overrightarrow{a}\cdot \overrightarrow{a}+\overrightarrow{b} \times \overrightarrow{c}\cdot \overrightarrow{a} + \overrightarrow{b}\times \overrightarrow{a}\cdot \overrightarrow{a}$
$=0+0+ \overrightarrow{b}\times \overrightarrow{a}\cdot \overrightarrow{c}+\overrightarrow{c} \times \overrightarrow{a}\cdot \overrightarrow{b}+0 + 0+0+\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a} + 0$
$= \overrightarrow{b}\times \overrightarrow{a}\cdot \overrightarrow{c}+\overrightarrow{c} \times \overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a}$