Let a and b be two vectors. Let |a|=1,|b|=4 and a b=2. If c=(2 a … - Vidyadip

MCQ

Let $\vec{a}$ and $\vec{b}$ be two vectors. Let $|\vec{a}|=1,|\vec{b}|=4$ and $\vec{a} \cdot \vec{b}=2$. If $\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}$, then the value of $\overrightarrow{ b } \cdot \overrightarrow{ c }$ is

A
$-24$
✓
$-48$
C
$-84$
D
$-60$

✓

Answer

Correct option: B.

$-48$

b
$\overrightarrow{ c }=(2 \overrightarrow{ a } \times \overrightarrow{ b })-3 \overrightarrow{ b }$

$\overrightarrow{ b } \cdot \overrightarrow{ c }=\overrightarrow{ b } \cdot(2 \overrightarrow{ a } \times \overrightarrow{ b })-3 \overrightarrow{ b } \cdot \overrightarrow{ b }$

$=-3| b |^2$

$=-48$

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 STD 12 - 10. vector algebra→STD 12 - 10. vector algebra — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

The value of $\int_{}^{} {\left( {1 + \frac{1}{{{x^2}}}} \right)\;{e^{\left( {x - \frac{1}{x}} \right)}}} \;dx$ equals

If $y = at^2 + 2bt + c$ and $t = ax^2 + 2bx + c$, then $\frac{{{d^3}y}}{{d{x^3}}}$ equals

Let $f: R_{+} \rightarrow[-5, \infty)$ be defined as $f(x)=9 x^2+6 x-5$, where $R_{+}$is the set of all non-negative real numbers. Then, $f$ is :

Solution set of the inequality $x \geq 0$ is

Choose the correct answer from the given four option.
The order and degree of the differential equation $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^{\frac{1}{4}}+\text{x}^{\frac{1}{5}}$ respectively, are:

If $y = {x^{\sqrt x }},$then ${{dy} \over {dx}} =$

The value of $\int_\pi ^{2\pi } {[2\sin x]\,dx,} $ where $[\,\,.\,\,]$ represents the greatest integer function, is

The value of f(0), so that the function $\text{f(x)}=\frac{(27-2\text{x})^\frac{1}{3}-3}{9-3(243+5\text{x})^\frac{1}{5}}$ is continuous, is given by:

If ${a_1},{a_2},{a_3}.....{a_n}....$ are in $G.P.$ then the value of the determinant $\left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{\log {a_{n + 1}}}&{\log {a_{n + 2}}}\\{\log {a_{n + 3}}}&{\log {a_{n + 4}}}&{\log {a_{n + 5}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 7}}}&{\log {a_{n + 8}}}\end{array}\,} \right|$ is

Area lying between the curves $y^2=4 x$ and $y=2 x$ is :