Let a and b be positive real numbers. Suppose | vedclass

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

Question

(D) The projection vector of $\vec{w}$ along $\vec{PQ}$ is $\vec{u} = \left( \frac{\vec{w} \cdot \vec{PQ}}{|\vec{PQ}|^2} \right) \vec{PQ}$.The magnitu

Vedclass · Accepted Answer

$A, C$

Vedclass · Answer

(D) The projection vector of $\vec{w}$ along $\vec{PQ}$ is $\vec{u} = \left( \frac{\vec{w} \cdot \vec{PQ}}{|\vec{PQ}|^2} ight) \vec{PQ}$.The magnitude is $|\vec{u}| = \frac{|\vec{w} \cdot \vec{PQ}|}{|\vec{PQ}|} = \frac{|(i+j) \cdot (ai+bj)|}{\sqrt{a^2+b^2}} = \frac{a+b}{\sqrt{a^2+b^2}}$.Similarly,the magnitude of the projection vector $\vec{v}$ along $\vec{PS}$ is $|\vec{v}| = \frac{|\vec{w} \cdot \vec{PS}|}{|\vec{PS}|} = \frac{|(i+j) \cdot (ai-bj)|}{\sqrt{a^2+b^2}} = \frac{|a-b|}{\sqrt{a^2+b^2}}$.Given $|\vec{u}| + |\vec{v}| = |\vec{w}| = \sqrt{1^2+1^2} = \sqrt{2}$,we have $\frac{a+b + |a-b|}{\sqrt{a^2+b^2}} = \sqrt{2}$.If $a \ge b$,then $\frac{2a}{\sqrt{a^2+b^2}} = \sqrt{2} \implies 4a^2 = 2(a^2+b^2) \implies 2a^2 = 2b^2 \implies a=b$.If $b > a$,then $\frac{2b}{\sqrt{a^2+b^2}} = \sqrt{2} \implies a=b$.Thus,$a=b$. The area of the parallelogram is $|\vec{PQ} 	imes \vec{PS}| = |(ai+bj) 	imes (ai-bj)| = |(-abk - abk)| = |-2abk| = 2ab = 8$.Since $a=b$,$2a^2 = 8 \implies a^2 = 4 \implies a=2$ (as $a>0$). Thus $a=2, b=2$.$(A)$ $a+b = 2+2 = 4$ (True).$(B)$ $a-b = 2-2 = 0 
eq 2$ (False).$(C)$ The diagonal $\vec{PR} = \vec{PQ} + \vec{PS} = (ai+bj) + (ai-bj) = 2ai = 4i$. Length is $|4i| = 4$ (True).$(D)$ $\vec{PQ} = 2i+2j$ and $\vec{PS} = 2i-2j$. The angle bisector direction is $\frac{\vec{PQ}}{|\vec{PQ}|} + \frac{\vec{PS}}{|\vec{PS}|} = \frac{2i+2j}{2\sqrt{2}} + \frac{2i-2j}{2\sqrt{2}} = \frac{4i}{2\sqrt{2}} = \sqrt{2}i$. This is along the $x$-axis,while $\vec{w} = i+j$ is at $45^\circ$. (False).

Similar Questions

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be any three non-coplanar vectors. If $m$ and $n$ are scalars such that $\vec{a}+\vec{b}=m \vec{d}-\vec{c}$ and $\vec{b}+\vec{c}=n \vec{a}-\vec{d}$,then $3 \vec{a}+2 \vec{b}+2 \vec{c}+\vec{d}=$

The position vectors of coplanar points $A, B, C,$ and $D$ are $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$ respectively,such that $(\vec{a} - \vec{d}) \cdot (\vec{b} - \vec{c}) = 0$ and $(\vec{b} - \vec{d}) \cdot (\vec{c} - \vec{a}) = 0.$ Then the point $D$ of the triangle $ABC$ is

The position vectors of the vertices of a quadrilateral $ABCD$ are $a, b, c$ and $d$ respectively. The area of the quadrilateral formed by joining the midpoints of its sides is

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that $\vec{b}$ and $\vec{c}$ are non-collinear. If $\vec{a}+5\vec{b}$ is collinear with $\vec{c}$,$\vec{b}+6\vec{c}$ is collinear with $\vec{a}$,and $\vec{a}+\alpha\vec{b}+\beta\vec{c}=\vec{0}$,then $\alpha+\beta$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module