Question
For any three vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ the expression $\big(\vec{\text{a}}-\vec{\text{b}}\big).\big\{\big(\vec{\text{b}}-\vec{\text{c}}\big)\times\big(\vec{\text{c}}-\vec{\text{a}}\big)\big\}$ equals:
  1. $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
  2. $2\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
  3. $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}]}^2$
  4. None of these

Answer

  1. None of these

Solution:

We have

$\big(\vec{\text{a}}-\vec{\text{b}}\big).\big[\big(\vec{\text{b}}-\vec{\text{c}}\big)\times\big(\vec{\text{c}}-\vec{\text{a}}\big)\big]$

$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big[\big(\vec{\text{b}}-\vec{\text{c}}\big)\times\vec{\text{c}}-\big(\vec{\text{b}}-\vec{\text{c}}\big)\times\vec{\text{a}}\big]$

$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{c}}-\vec{\text{c}}\times\vec{\text{c}}-\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}\big)$

$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times{\vec{\text{c}}}-0-\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}\big)$

$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{a}}\big)+\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{c}}\times\vec{\text{a}}\big)$

$=\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\vec{\text{b}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\vec{\text{a}}\big(\vec{\text{b}}\times\vec{\text{a}}\big)\\+\vec{\text{b}}.\big(\vec{\text{b}}\times\vec{\text{a}}\big)+\vec{\text{a}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)-\vec{\text{b}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)$

$=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]-0-0+0+0-\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]$ $\big(\therefore\big[\vec{\text{b}}\vec{\text{b}}\vec{\text{c}}\big]=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\big]=\big[\vec{\text{b}}\vec{\text{b}}\vec{\text{a}}\big]=0\big)$

$=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$ $\big(\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]=\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]\big)$

$=0$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Let $y = g (x)$ be the inverse of a bijective mapping $f : R \rightarrow R \, f (x) = 3x^3 + 2x. $ The area bounded by the graph of $g (x),$ the $x-$ axis and the ordinate at $x = 5$ is :
Let, $f(x)=\left\{\begin{array}{l} x \sin \left(\frac{1}{x}\right) \text { when } x \neq 0 \\ 1 \text { when } x=0 \end{array}\right\}$ and $A=\{x \in R: f(x)=1\} .$ Then, $A$ has
The minimum value of $f(x) = |x - 1| + |2x - 1| + |3x - 1| + ...... + |119x - 1|$ occurs at $x$. Then $x$ is equal to
The relation $R$ defined in $N$ as $aRb \Leftrightarrow b$ is divisible by $a$ is
Let $(\alpha, \beta, \gamma)$ be the foot of perpendicular from the point $(1,2,3)$ on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$. then $19(\alpha+\beta+\gamma)$ is equal to :
If $f(x) = |x - 2|$, then
If for $p \neq q \neq 0$, then function,$f(x)=\frac{\sqrt[7]{p(729+x)}-3}{\sqrt[3]{729+q x}-9}$is continuous at $x=0$, then
The points represented  by $\vec a,\vec b,\vec c,\vec d$ are coplanar and $\left( {\sin A} \right)\vec a + \left( {2\sin 2B} \right)\vec b + \left( {3\sin 3C} \right)\vec c - 4\vec d = \vec 0$ then the least value of $\frac{{21}}{8}\left( {{{\sin }^2}A + {{\sin }^2}2B + {{\sin }^2}3C} \right)$ is
The angle made by line $\text{r}[\cos\theta−3​\sin\theta]=5 $ with initial line is:
Consider the binary operation * defined on Q − {1} by the rule a * b = a + b − ab for all a, b ∈ Q − {1}. The identity element in Q − {1} is:
  1. $0$
  2. $1$
  3. $\frac{1}2$
  4. $-1$