Question

If vectors 2i^+2j^-2k^,5i^+yi^+k^ are perpendicular to each other. The value of 'y'is

Answer 1

$\huge\boxed{\fcolorbox{cyan}{grey}{Solution:-}}$.

➡Let, Given vectors are ,

♠ â= 2î+2j-2k .

♠b=5î+yj+k.

➡They are perpendicular .,

so, by conditions.

$\huge\boxed{\fcolorbox{cyan}{grey}{</u></em></strong><strong><em><u>â</u></em></strong><strong><em><u>.</u></em></strong><strong><em><u>b</u></em></strong><strong><em><u>=</u></em></strong><strong><em><u>0</u></em></strong><strong><em><u>:-}}$.

Now,.

=> (2î+2j-2k).(5î+yj+k)=0.

=>10+2y-2=0.

=>2y= -8.

=>y= -4.

➡Hopes its helps u.

$\huge\boxed{\fcolorbox{cyan}{grey}{</u></em></strong><strong><em><u>F</u></em></strong><strong><em><u>o</u></em></strong><strong><em><u>l</u></em></strong><strong><em><u>l</u></em></strong><strong><em><u>o</u></em></strong><strong><em><u>w</u></em></strong><strong><em><u> </u></em></strong><strong><em><u>m</u></em></strong><strong><em><u>e</u></em></strong><strong><em><u>.</u></em></strong><strong><em><u>:-}}$

Answer 2

Answer:

$\large\bold\red{y=-4}$

Explanation:

Given,

Two vectors,

Let's assume that,

• $\bold{\vec{a}=2\hat{i}+2\hat{j}-2\hat{k}}$

And,

• $\bold{\vec{b}=5\hat{i}+y\hat{j}+\hat{k}}$

Also,

It's given that,

• Both vectors are perpendicular to each other.

Now,

We know that,

• Dot Product of Perpendicular vectors is zero.

Therefore,

We get,

$= > \vec{a} \cdot\vec{b} = 0 \\ \\ = > (2\hat{i}+2\hat{j}-2\hat{k}) \cdot(5\hat{i}+y\hat{j}+\hat{k}) = 0$

But,

We know that,

• Dot product of Unit vector with itself is 1 and with another unit vector is 0.

Therefore,

We get,

$= > 10 + 2y - 2 = 0 \\ \\ = > 2y + 8 = 0 \\ \\ = > 2y = - 8 \\ \\ = > y = \frac{ - 8}{2} \\ \\ = > y = - 4$

Hence,

• $\large\bold{y=-4}$