Let a and b be two unit vectors such that the angle between them is ( )/(3).

Question

Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that the angle between them is $\frac{\pi }{3}$. If $\lambda \vec{a}+2\vec{b}$ and $3\vec{a}-\lambda \vec{b}$ are perpendicular to each other, then the number of values of $\lambda $ in $[-1,3]$ is :

Accepted Answer

We are given two unit vectors $\vec{a}$ and $\vec{b}$ with an angle of $\frac{\pi }{3}$ between them. This means:

$\vec{a}\cdot \vec{a}=\vec{b}\cdot \vec{b}=1$ (since they are unit vectors)

$\vec{a}\cdot \vec{b}=\cos \frac{\pi }{3}=\frac{1}{2}$

We need to find the number of values of $\lambda $ in the interval $[-1,3]$ for which the vectors $\lambda \vec{a}+2\vec{b}$ and $3\vec{a}-\lambda \vec{b}$ are perpendicular, meaning their dot product is zero.

Here's how to solve the problem step by step:

Set up the perpendicular condition by equating the dot product to zero:

$(\lambda \vec{a}+2\vec{b})\cdot (3\vec{a}-\lambda \vec{b})=0.$

Expand the dot product using the distributive property:

$\begin{gathered} (\lambda \vec{a}+2\vec{b})\cdot (3\vec{a}-\lambda \vec{b}) =\lambda \cdot 3(\vec{a}\cdot \vec{a})-{\lambda }^{2}(\vec{a}\cdot \vec{b})+2\cdot 3(\vec{b}\cdot \vec{a})-2\lambda (\vec{b}\cdot \vec{b}) \ =3\lambda -{\lambda }^{2}(\frac{1}{2})+6(\frac{1}{2})-2\lambda \ =3\lambda -\frac{{\lambda }^{2}}{2}+3-2\lambda . \end{gathered}$

Simplify the expression:

$3\lambda -2\lambda =\lambda ,$

so the equation becomes:

$\lambda -\frac{{\lambda }^{2}}{2}+3=0.$

Multiply the entire equation by 2 to eliminate the fraction:

$2\lambda -{\lambda }^{2}+6=0,$

which can be rearranged as:

$-{\lambda }^{2}+2\lambda +6=0.$

Multiply both sides by -1 to get a standard quadratic form:

${\lambda }^{2}-2\lambda -6=0.$

Solve the quadratic equation using the quadratic formula:

$\lambda =\frac{2\pm \sqrt{(-2{)}^{2}-4\cdot 1\cdot (-6)}}{2}=\frac{2\pm \sqrt{4+24}}{2}=\frac{2\pm \sqrt{28}}{2}.$

Notice that:

$\sqrt{28}=2\sqrt{7},$

thus:

$\lambda =\frac{2\pm 2\sqrt{7}}{2}=1\pm \sqrt{7}.$

Evaluate the solutions numerically:

$\lambda =1+\sqrt{7}\approx 1+2.6458\approx 3.6458,$

$\lambda =1-\sqrt{7}\approx 1-2.6458\approx -1.6458.$

Check if the solutions are in the interval $[-1,3]$:

$1+\sqrt{7}\approx 3.6458$ is greater than 3.

$1-\sqrt{7}\approx -1.6458$ is less than -1.

Since neither value lies within the interval $[-1,3]$, there are no valid values of $\lambda $ in that interval.

Therefore, the number of values of $\lambda $ in the interval $[-1,3]$ is:

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