Download App

VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA1 Mark
MCQ
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:
  • ✓
    0
  • B
    2
  • C
    1
  • D
    None of these
✓

Answer

Correct option: A.
0
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→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If $(a, 0); a > 0$ is the point where the curve $y = \sin 2x - \sqrt 3 \, \sin x$ cuts the $x-$ axis first, $A$ is the area bounded by this part of the curve , the origin and the positive $x-$ axis, then
→
The area of the region bounded by the line $y = | x - 2 |, x = 1, x = 3$ and $x-$axis is:
→
Let $a$ and $b$ be positive real numbers. Suppose $\overrightarrow{ PQ }=a \hat{ i }+b \hat{ j }$ and $\overrightarrow{ PS }=a \hat{ i }-b \hat{ j }$ are adjacent sides of $a$ parallelogram $P Q R S$. Let $\overrightarrow{ u }$ and $\overrightarrow{ v }$ be the projection vectors of $\overrightarrow{ w }=\hat{ i }+\hat{ j }$ along $\overrightarrow{ PQ }$ and $\overrightarrow{ PS }$, respectively. If $|\vec{u}|+|\vec{v}|=|\vec{w}|$ and if the area of the parallelogram $P Q R S$ is $8$ , then which of the following statements is/are $TRUE$?

$(A)$ $a+b=4$

$(B)$ $a-b=2$

$(C)$ The length of the diagonal $P R$ of the parallelogram $P Q R S$ is $4$

$(D)$ $\overrightarrow{ w }$ is an angle bisector of the vectors $\overrightarrow{ PQ }$ and $\overrightarrow{ PS }$

→
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:
→
The inverse of the matrix $\left[ {\begin{array}{*{20}{c}}3&{ - 2}\\1&4\end{array}} \right]$is
→
The determinant $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha + b}\\b&c&{b\alpha + c}\\{a\alpha + b}&{b\alpha + c}&0\end{array}\,} \right| = 0$, if $a,b,c$ are in
→
if $\text{ax}+\frac{\text{b}}{\text{x}}\geq\text{c}$ for all positive x where a, b, > 0, then.
→
If $f(x) = \left\{ \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,1,\,\,x < 0\\1 + \sin x,\,\,0 \le x < \frac{\pi }{2}\end{array} \right.$ then $f'(0) = $
→
A normal to the plane x = 2 is:
→
Let $\text{f(x)}=|\cos\text{x}|.$ Then,
→