Question

How to solve this question ? o.O
If \[4\mathbf{i}-3\mathbf{j}\] and \[\lambda\mathbf{i}+2\mathbf{j}\] are perpendicular to each other, find the value of \[\lambda\].

Answer 1

hint : dot product equals 0 when vectors are \(\perp\)

Answer 2

\((4i - 3j) . (\lambda i + 2j) = 0 \)

Answer 3

\(4\lambda - 6 = 0 \)

Answer 4

But how did it become \[4\lambda-6=0\]

Answer 5

good question :)
lets work it step by step :-
FOIL it first

Answer 6

I am very bad at vectors. :( What do you mean by FOIL?

Answer 7

FOIL is just fancy term lol. just multiply the product

Answer 8

Ohh :)

Answer 9

\((4i - 3j) . (\lambda i + 2j) = 0 \)

\(4i . (\lambda i + 2j) - 3j . (\lambda i + 2j) \)

\(4i . (\lambda i) + 4i . (2j) - 3j . (\lambda i) - 3j . ( 2j) \)

\(4\lambda (i.i) + 8(i . j) - 3\lambda (j.i) - 6 (j.j) \)

Answer 10

see if t

Answer 11

Yeah got it.

Answer 12

makes sense so far.

Answer 13

i and j becomes 0

Answer 14

i and i becomes 1

Answer 15

next use below properties :-
\(i . i = j . j = 1\)
\(i . j = j . i = 0\)

Answer 16

yos you got it !

Answer 17

btw, \(i.j\) becomes zero, becoz by definition \(i \perp j\)

Answer 18

So the answer becomes \[\lambda=\frac{3}{2}\]. That's the answer my teacher gave too. :)

Answer 19

\[\mathbf{i}.\mathbf{j}=|\mathbf{i}|.|\mathbf{j}| cos 90\]
\[\mathbf{i}.\mathbf{j}=|\mathbf{i}|.|\mathbf{j}| \times0\]
\[\mathbf{i}.\mathbf{j}=0\]

Answer 20

Hope I am correct. Thanks for the medal too. :)

Answer 21

Exactly !

Answer 22

\(4i - 3j\)
\(\lambda i + 2j\)

short cut for taking dot product of above two vectors is to simply multiply \(i\) components and \(j\) components separately and add them :-

product of \(i\) components :-
\(4(\lambda) \)

product of \(j\) components :-
\(-3(2)\)


add them both :-
\(4 \lambda - 6\)

Answer 23

using ur method u can also try proving why \(i . i = 1\) :)