NEET Physics Mathematical Tools Questions Solved

(1)

$$\begin{gathered} \mathrm{Vector} \overrightarrow{\mathrm{A}} \mathrm{has} 2 \mathrm{components}, \\ \mathrm{Component} \mathrm{of} \overrightarrow{\mathrm{A}} \mathrm{along} \overrightarrow{\mathrm{B}} ,{\overrightarrow{\mathrm{A}}}_{\parallel } =\frac{\left(\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}\right)\overrightarrow{\mathrm{B}}}{{\mathrm{B}}^2} \\ \mathrm{Component} \perp \mathrm{to} \overrightarrow{\mathrm{B}}={\overrightarrow{\mathrm{A}}}_{\perp } \\ \mathrm{So}, \overrightarrow{\mathrm{A}}={\overrightarrow{\mathrm{A}}}_{\parallel }+{\overrightarrow{\mathrm{A}}}_{\perp } \\ {\overrightarrow{\mathrm{A}}}_{\perp }=\overrightarrow{\mathrm{A}}-{\overrightarrow{\mathrm{A}}}_{\parallel } \\ = \overrightarrow{\mathrm{A}}-\frac{\left(\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}\right)\overrightarrow{\mathrm{B}}}{{\mathrm{B}}^2} \end{gathered}$$

$$\begin{gathered} \mathrm{Perpendicular} \mathrm{component} \mathrm{of} \overrightarrow{\mathrm{A}} \mathrm{to} \overrightarrow{\mathrm{B}} \mathrm{is} \\ = \overrightarrow{\mathrm{A}}-\frac{\left(\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}\right)\overrightarrow{\mathrm{B}}}{{\left|\mathrm{B}\right|}^2}= \left(3\hat{\mathrm{i}}+4\hat{\mathrm{j}}\right)-\frac{7}{2}\left(\mathrm{i}+\hat{\mathrm{j}}\right) \\ = \frac{-\mathrm{i}}{2}+\frac{\mathrm{j}}{2} \end{gathered}$$

Using sin rule, and taking case where $\alpha < \beta$

(2)

Let angle betwee A & B is $\mathrm{\theta }$

$$\begin{gathered} {\mathrm{R}}_1^2={\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{ABcos\theta } \\ \tan 90°=\frac{1}{0}=\frac{2\mathrm{Bsin\theta }}{\mathrm{A}+2\mathrm{Bcos\theta }} \\ \mathrm{A}=-2\mathrm{Bcos\theta } \\ {\mathrm{R}}_1^2=4{\mathrm{B}}^2{\cos }^2\mathrm{\theta }+{\mathrm{B}}^2-4{\mathrm{B}}^2{\cos }^2\mathrm{\theta } \\ {\mathrm{R}}_1=\mathrm{B} \end{gathered}$$

(2) $A+B=18$ …(i)

$12=\sqrt{A^2+B^2+2AB\cos \theta }$ …(ii)

$\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }=\tan 90°$

⇒ $\cos \theta =-\frac{A}{B}$ …(iii)

By solving (i), (ii) and (iii),

A = 13 N and B = 5 N

(a) Suppose two vectors are P and Q. It is given that [P+Q]=[P-Q]

Let angle between P and Q is φ.
∴  P²+Q²+2PQ cosφ=P²+Q²-2PQcosφ
=> 4PQcosφ=0
=> cosφ=0   [∴ P,Q≠0]
=> φ=π/2=90° 

If two non-zero vectors are represented by the two adjust sides of a parallelogram, then the resultant is given by the diagonal of the parallelogram passing through the point of intersection of the two vectors 

$\therefore \overrightarrow{b}+\overrightarrow{c}=\overrightarrow{-f}$

(C)

Angle between vectors is angle between tails/heads.

For vector $\overrightarrow{\mathrm{A}} \& \overrightarrow{\mathrm{B}}$,

we can translate vector $\overrightarrow{\mathrm{A}}$

Now tails are joined  $\mathrm{\theta }= 70°$

For vector $\overrightarrow{\mathrm{B}} \& \overrightarrow{\mathrm{C}}$,

 we can translate vector $\overrightarrow{\mathrm{B}}$

Tails are joined  $\mathrm{\theta }= 110°$

(B)   

For 2 vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$,

Depending on $\mathrm{\theta }$, magnitude of resultant will lie between-

                $\left|\mathrm{A}-\mathrm{B}\right|<\mathrm{R}<\mathrm{A}+\mathrm{B}$

In question,

             $8-6 <\mathrm{R}<8+6$

             $2\mathrm{N}<\mathrm{R}<14\mathrm{N}$

(C)

$$\begin{gathered} \mathrm{Let} \mathrm{there} \mathrm{be} 3 \mathrm{vectors} \overrightarrow{\mathrm{A}}, \overrightarrow{\mathrm{B}} \& \overrightarrow{\mathrm{C}}. \\ \mathrm{Suppose} \mathrm{magnitude} \mathrm{of} \mathrm{C} \mathrm{is} \mathrm{highest}. \\ \mathrm{For} \mathrm{zero} \mathrm{resultant}, \\ \mathrm{C} \le \mathrm{A}+\mathrm{B} \\ \mathrm{Where} \mathrm{A}, \mathrm{B} \& \mathrm{C} \mathrm{represents} \mathrm{magnitude} \mathrm{of} \overrightarrow{\mathrm{A}}, \overrightarrow{\mathrm{B}} \& \overrightarrow{\mathrm{C}} \\ \mathrm{For} \mathrm{pair} 1, 2, 1 \\ \mathrm{C}= 2 \\ \mathrm{A}= 1 \\ \mathrm{B}= 1 \\ \mathrm{Also}, 2 \le 1+1 \end{gathered}$$