If ${{\bf{\vec a}}}$ and ${{\bf{\vec b}}}$ are unit vectors and $\left| {{\bf{\vec a}} + {\bf{\vec b}}} \right| = 1$, then $\left| {{\bf{\vec a}} + {\bf{\vec b}}} \right|$ is equal to

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Application +4 | -1

{\mathbf{\vec a}}

and

{\mathbf{\vec b}}

are unit vectors and

\left| {\mathbf{\vec a} + \mathbf{\vec b}} \right| = 1

, then

\left| {\mathbf{\vec a} + \mathbf{\vec b}} \right|

is equal to

View answer and explanation

Correct option: D

\sqrt 3

Option D (

\sqrt 3

) is correct because using the given conditions and the properties of dot products, the magnitude of the difference vector is found to be

\sqrt{3}

Practice All Chapter MCQs →

\vec r\cdot\vec a = \vec r\cdot\vec b = \vec r\cdot\vec c = 0

\vec r,

\left[ {\vec a\vec b\vec c} \right],

\frac{{x - 3}}{2} = \frac{{y - 2}}{3} = \frac{{z - 1}}{{ - 1}}

\frac{{x + 3}}{2} = \frac{{y - 6}}{1} = \frac{{z - 5}}{3},

\mathbf{\vec a} = \widehat \mathbf{i} - \mathbf{\hat k},\;\mathbf{\vec b} = x\widehat \mathbf{i} + \widehat \mathbf{j} + \left( {1 - x} \right)\mathbf{\hat k}

\mathbf{\vec c} = y\widehat \mathbf{i} + x\widehat \mathbf{j} + \left( {1 + x - y} \right)\mathbf{\hat k}.

\left[ {\mathbf{\vec a},\;\mathbf{\vec b},\;\mathbf{\vec c}} \right]

\vec{a}, \vec{b}

((\vec{a} \times \vec{b}) \times \vec{a}) \cdot((\vec{b} \times \vec{a}) \times \vec{b})

a \cdot(a \times b)=

\mathbf{r}=\mathbf{b}+\lambda \mathbf{c}