Download App

VECTOR ALGEBRA — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSVECTOR ALGEBRA1 Mark
Question
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are non-coplanar vectors, then $\frac{\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)}{\big(\vec{\text{c}}\times\vec{\text{a}}\big).\vec{\text{b}}}+\frac{\vec{\text{b}}.\big(\vec{\text{a}}\times\vec{\text{c}}\big)}{\vec{\text{c}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)}$ is equal to:
  1. 0
  2. 2
  3. 1
  4. None of these
✓

Answer

  1. 0
Solution:
We have
$\frac{\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)}{\big(\vec{\text{c}}\times\vec{\text{a}}\big).\vec{\text{b}}}+\frac{\vec{\text{b}}.\big(\vec{\text{a}}\times\vec{\text{c}}\big)}{\vec{\text{c}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)}$
$\frac{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}{\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]}+\frac{\big[\vec{\text{b}}\vec{\text{a}}\vec{\text{c}}\big]}{\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]}$ (By definition of scalar tiple product)
$=\frac{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}+\frac{-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}$ (Change in cyclic order of vectors changes the sign of the scalar triple product)
$=1-1$
$=0$

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

Explore more

More "M.C.Q (1 Marks)" from VECTOR ALGEBRA→VECTOR ALGEBRA — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Let $\sin\text{y}=\text{x}\sin(\text{a}+\text{y}),$ then $\frac{\text{dy}}{\text{dx}}$ is:
  1. $\frac{\sin\text{a}}{\sin\text{a}\sin^2(\text{a}+\text{y})}$
  2. $\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
  3. $\sin\text{a}\sin^2(\text{a}+\text{y})$
  4. $\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
→
If $\left[\begin{array}{cc}x & 2 \\ 18 & x\end{array}\right]=\left[\begin{array}{cc}6 & 2 \\ 18 & 6\end{array}\right]$ then $x$ is equal to :
→
If function $f(x)=\left\{1+x, x \leq 2k-\frac{x}{2}, x>2\right.$, is continuous at $x=2$, then value of $k$ will be :
→
The function $f(x)=\frac{4-x^2}{4 x-x^3}$ is
→
If $6 \sin ^{-1}\left(x^2-6 x+8.5\right)=\pi$, then $x$ is equal to
→
Evaluate: $\int\left(e^{x \log a}+e^{a \log x}+e^{a \log a}\right) d x$
→
The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both the points (15, 15) and (0, 20) is Maximum of Z occurs at:
  1. (5, 0)
  2. (6, 5)
  3. (6, 8)
  4. (4, 10)
→
If $\sin^{-1}\text{x}-\cos^{-1}\text{x}=\frac{\pi}{6},$ then x =
  1. $\frac{1}{2}$
  2. $\frac{\sqrt{3}}{2}$
  3. $-\frac{1}{2}$
  4. $-\frac{\sqrt{3}}{2}$
→
Region represented by $\text{x}\geq0, \text{y}\geq0$ is:
  1. First quadrant
  2. Second quadrant
  3. Third quadrant
  4. Fourth quadrant
→
The straight line $\frac{\text{x}-3}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-1}{0}$ is:
→