Have you practised your Vectors?

Geometry Level 2

If $\vec A$ and $\vec B$ are two vectors such that $\vec A + \vec B$ is perpendicular to $\vec B$ and $\vec A+2\vec B$ is perpendicular to $\vec A$ , which of the following is correct?

Clarification: The vectors $\vec A$ and $\vec B$ are both non-zero.

2\left \| \vec A\right \|=\left \| \vec B\right \|

\left \| \vec A\right \|=\left \| \vec B\right \|

\left \| \vec A\right \|=\sqrt 2 \left \| \vec B\right \|

\left \| \vec A\right \| = 2\left \| \vec B\right \|

2 solutions

Sravanth C.
May 21, 2016

We know that if two vectors are perpendicular, their dot product is $0$ . Thus we have;

$\begin{aligned}(\vec A + \vec B)\cdot \vec B &= 0\\ \vec A \cdot \vec B + \vec B \cdot \vec B & =0\\ \vec A \cdot\vec B &= -|\vec B|^2 \end{aligned}$

Similarly we have;

$\begin{aligned}(\vec A + 2\vec B)\cdot \vec A &= 0\\ \vec A \cdot \vec A + 2\vec B \cdot \vec A & =0\\ 2\vec B \cdot\vec A &= -\vec |A|^2 \end{aligned}$

Now, combining both the results, we get;

$\begin{aligned} \dfrac{2\vec B\cdot \vec A}{\vec A\cdot\vec B}&=\dfrac{|\vec A|^2}{|\vec B|^2}\\ \sqrt 2|\vec B| & = |\vec A|\\ \end{aligned}$

Moderator note:

You found a necessary condition for these 2 vectors. Do non-zero vectors that satisfy the conditions exist?

Nice solution +1 bro.

I did it by using the formula for angle b/w the resultant & any vector using tan¤ !! I applied it to both cases & then Equated the results ! Nice ques.! Btw, in which class u r?

Rishabh Tiwari - 5 years ago

Ahah thanks! I am in 11th presently.

Sravanth C. - 5 years ago

Ok me 2 ! We r even in the same institution of kendriya vidyalaya.. but ur really much better than me bro.! U might have definitely rocked in your board exams!! Have u been using brilliant from 10th. Class?

Rishabh Tiwari - 5 years ago

@Rishabh Tiwari – Oh nice you are from KV. Well the boards were easy hope the result is will be good. Yeah I've been using brilliant from 10th. How was your boards?

Sravanth C. - 5 years ago

FYI The conclusion is about the magnitude, and not the vector directions. I've edited the options to reflect that.

Calvin Lin Staff - 5 years ago

Thanks sir, I was not able to put the modulus sign.

Sravanth C. - 5 years ago

Yea, we prevent using "|" in the answer choices for technical reasons, so I used "\vert" instead.

Calvin Lin Staff - 5 years ago

Note: You should mention that the vectors are non-zero, in order to perform the division.

Calvin Lin Staff - 5 years ago

@Calvin Lin – I don't think that's necessary; the result is still true if one of the vectors is zero, and his proof still works: it's always ok to take square roots of $|A|^2 = 2|B|^2.$

Patrick Corn - 5 years ago

@Patrick Corn – I agree that the result is still true. My conditional was "in order to perform the division". If the "combining both results" step was written instead as $2|B|^2 = -2 A \cdot B = |A|^2$ , then we didn't need to divide by $|B|^2$ as in the proof.

Of course, if we wanted to do the division, we can take $|B| = 0$ as an edge case, and then figure out what happens in that scenario. As it turns out $A \perp A$ which implies that $A=0$ and hence the equation still applies.

Calvin Lin Staff - 5 years ago

An example satisfying the conditions: A = (-1, 1) and B = (1, 0). So A + B = (0, 1) and A + 2B = (1, 1)

Ali Hadizadeh - 2 years, 5 months ago

Jit Sadhu
May 4, 2021

From Cosine rule and Pythagoras theorem,

$(|A + B|)^2 = |A|^2 + |B|^2 + 2|A||B| \cosθ = |A|^2 - |B|^2$

So, $|B| = - |A| \cosθ$ --(1)

$(|A + 2B|)^2 = |A|^2 + |2B|^2 + 2|A||2B| \cosθ = |2B|^2 - |A|^2$

So, $|A| = -2|B| \cos θ$

or, $|A| = 2|A| \cos^2 θ$ [From (1)]

or, $\cos θ = ± \dfrac1{\sqrt 2}$

Hence, $|A| = \sqrt 2 |B|$ (as magnitude is positive)

Have you practised your Vectors?

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