Download App

10. vector algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths10. vector algebra1 Mark
MCQ
$|a \times i{|^2} + |a \times j{|^2} + |a \times k{|^2} = $
  • A
    $|a{|^2}$
  • ✓
    $2\,\,|a{|^2}$
  • C
    $3\,\,|a{|^2}$
  • D
    $4\,\,|a{|^2}$
✓

Answer

Correct option: B.
$2\,\,|a{|^2}$
b
(b) $|a \times i{|^2} = {\left| {\begin{array}{*{20}{c}}i&j&k\\{{a_1}}&{{a_2}}&{{a_3}}\\1&0&0\end{array}} \right|^2}$, $({\rm{Since}}\,\,\,a = {a_1}i + {a_2}j + {a_3}k)$

$ = \,|{a_3}j - {a_2}k{|^2} = a_3^2 + a_2^2$

Similarly, $|a \times j{|^2} = a_1^2 + a_3^2$ and $|a \times k{|^2} = a_1^2 + a_2^2$

Hence the required result can be given as

$2(a_1^2 + a_2^2 + a_3^2) = 2|a{|^2}.$

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

Explore more

More "M.C.Q (1 Marks)" from 10. vector algebra→10. vector algebra — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

$\int {{{\sin }^{\frac{{ - 1}}{2}}}x{{\cos }^{\frac{{ - 7}}{2}}}xdx = } $
→
A problem in Mathematics is given to three students whose chances of solving it are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ respectively. If the events of their solving the problem are independent then the probability that the problem will be solved, is
→

A coin is tossed n times. The probability of geting at least once is greater than 0.8. Then, the least value of n, is:

  1. 2
  2. 3
  3. 4
  4. 5
→
If in a right angled triangle $ABC$, the hypotenuse $AB = p,$ then $\overrightarrow {AB} \,\,.\,\,\overrightarrow {AC} + \overrightarrow {BC} \,.\,\,\overrightarrow {BA} + \overrightarrow {CA} \,\,.\,\,\overrightarrow {CB} $ is equal to
→
If $\text{y}=\frac{\text{ax}+\text{b}}{\text{x}^2+\text{c}},$ then $(2\text{xy}_1+\text{y})\text{y}_3=$
  1. 3(xy2 + y1) y2
  2. 3(xy1 + y2) y2
  3. 3(xy1 + y2) y1
  4. None of these
→
Let $g(x)=\log (f(x))$ where $f(x)$ is a twice differentiable positive function on $(0, \infty)$ such that $f(x+1)=x f(x)$. Then, for $\mathrm{N}=1,2,3, \ldots$,

$\mathrm{g}^{\prime \prime}\left(\mathrm{N}+\frac{1}{2}\right)-\mathrm{g}^{\prime \prime}\left(\frac{1}{2}\right)=$

→
Let $\int_\alpha^{\log _e^4} \frac{\mathrm{dx}}{\sqrt{\mathrm{e}^{\mathrm{x}}-1}}=\frac{\pi}{6}$. Then $\mathrm{e}^\alpha$ and $\mathrm{e}^{-\alpha}$ are the roots of the equation :
→
If ${\tan ^{ - 1}}x + {\cos ^{ - 1}}\frac{y}{{\sqrt {(1 + {y^2})} }} = {\sin ^{ - 1}}\frac{3}{{\sqrt {10} }}$ and both $x$ and $y$ are positive and integral, then $x$ and $y =$
→
If $\,\,f(x) = \left\{ {\begin{array}{*{20}{c}}
  {3 + x;\,\,\,\,\,x \geqslant 0} \\ 
  {2 - 3x;\,\,\,\,\,x < 0} 
\end{array}} \right.$ then $\mathop {\lim }\limits_{x \to 0} f(f(x))$ is equal to -
→
$\int_{}^{} {{e^x}(1 + \tan x)\sec x\;dx = } $
→