First slide

Dot product or scalar product of vectors

Question

$Let a and b be positive real numbers. Suppose \vec{PQ} = a \hat{i} + b \hat{j} and \vec{PS} = a \hat{i} - b \hat{j} are adjacent sides of a$
$parallelogram P Q R S . Let \vec{u} and \vec{v} be the projection vectors of \vec{w} = \hat{i} + \hat{j} along \vec{PQ} and \vec{PS},$
$respectively. If | \vec{u} | + | \vec{v} | = | \vec{w} | and if the area of the parallelogram P Q R S is 8, then which of the following statements is/are TRUE?$

Difficult

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

$\vec{w} is an angle bisector of the vectors \vec{PQ} and \vec{PS}$

Solution

$s i m i l a r l y, \begin{array}{l} \vec{u} = p r o j e c t i o n v e c t o r o f \hat{i} + \hat{j} o n P Q, h e n c e i t s l e n g t h = | \vec{u} | = |(i + j) \cdot \frac{(ai + bj)}{\sqrt{a^{2} + b^{2}}}| = \frac{a + b}{a^{2} + b^{2}} \\ | \vec{v} | = |\frac{(i + j) \cdot (ai - bj)}{\sqrt{a^{2} + b^{2}}}| = \frac{a - b}{\sqrt{a^{2} + b^{2}}} \\ |\vec{u}| + | \vec{v} | = | \vec{w} | \\ \frac{| (a + b) | + | (a - b) |}{\sqrt{a^{2} + b^{2}}} = \sqrt{2} \end{array}$
$\begin{array}{l} For a \geq b \\ 2 a = \sqrt{2} \cdot \sqrt{a^{2} + b^{2}} \\ 4 a^{2} = 2 a^{2} + 2 b^{2} \\ a^{2} = b^{2} ∴ a = b \dots . . . . (1) \end{array}$
$\begin{array}{l} (a > 0, b > 0) \\ similarly for b \geq a we will get a = b \\ Now area of parallelogram = | (ai + bj) \times (ai - bj) | \end{array}$
$\begin{array}{l} = 2 ab \\ ∴ 2 ab = 8 \\ ab = 4 . . . . . . . (2) \end{array}$
from (1) and (2)
$a = 2, b = 2 ∴ a + b = 4 option (A)$
length of diagonal is $| 2 ai | = | 4 \hat{i} | = 4$
so option (C)

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App