Let the vectors overline{PQ}, overline{QR}, o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In a regular hexagon $PQRSTU$,the sides are vectors. Let $\vec{a} = \overline{PQ}$. Then $\overline{RS}$ is parallel to $\overline{UP}$ and $\over

Vedclass · Accepted Answer

$Statement-1$ is True,$Statement-2$ is False.

Vedclass · Answer

(C) In a regular hexagon $PQRSTU$,the sides are vectors. Let $\vec{a} = \overline{PQ}$. Then $\overline{RS}$ is parallel to $\overline{UP}$ and $\overline{ST}$ is parallel to $\overline{PQ}$.Specifically,$\overline{RS} = -\overline{PQ} = -\vec{a}$ is incorrect; rather,$\overline{RS}$ is parallel to $\overline{PQ}$ but in the opposite direction,so $\overline{RS} = -\overline{PQ}$ is false. In a regular hexagon,$\overline{PQ} = \overline{UT}$ and $\overline{RS} = \overline{PQ}$ is not true. Actually,$\overline{PQ} \parallel \overline{ST}$ and $\overline{RS} \parallel \overline{UP}$.Since $\overline{PQ}$ is not parallel to $\overline{RS}$,$\overline{PQ} 	imes \overline{RS} 
eq \overrightarrow{0}$.Thus,$Statement-2$ is False because it claims $\overline{PQ} 	imes \overline{RS} = \overrightarrow{0}$.For $Statement-1$,$\overline{PQ} 	imes (\overline{RS} + \overline{ST}) = \overline{PQ} 	imes \overline{RS} + \overline{PQ} 	imes \overline{ST}$. Since $\overline{PQ}$ is not parallel to the resultant $\overline{RS} + \overline{ST}$,the cross product is non-zero. Thus,$Statement-1$ is True.

Similar Questions

If $\overline{a}=3 \hat{i}-5 \hat{j}$ and $\overline{b}=6 \hat{i}-3 \hat{j}$ are two vectors and $\overline{c}$ is a vector such that $\overline{c}=\overline{a} \times \overline{b}$,then the ratio $a: b: c$ is:

Let $p, q, r$ be three mutually perpendicular vectors of the same magnitude. If a vector $x$ satisfies the equation $p \times \{(x - q) \times p\} + q \times \{(x - r) \times q\} + r \times \{(x - p) \times r\} = 0$,then $x$ is given by

For a triangle $\Delta ABC$,if $\vec{BC} = \vec{a}$,$\vec{CA} = \vec{b}$,and $\vec{AB} = \vec{c}$,then which of the following is true?

The area of a triangle with vertices $(1, 2, 0)$,$(1, 0, a)$,and $(0, 3, 1)$ is $\sqrt{6}$ sq. units. Then the values of '$a$' are:

Let $a=2 \hat{i}-2 \hat{j}+\hat{k}$ and $b=-\hat{j}+\hat{k}$. If $c$ is a vector such that $a \cdot c=|c|$, $|c-a|=2 \sqrt{2}$, and the angle between $a \times b$ and $c$ is $\frac{\pi}{3}$, then $|(a \times b) \times c|=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module