Question

if n cap= aicap +bjcap is perpendicular to the vector (icap +jcap) then find the value of a&b may be​

Answer 1

Answer:

The value of a and b are $-\sqrt{\dfrac{1}{2}}, \sqrt{\dfrac{1}{2}}$

Explanation:

Given that,

$\hat{n}=a\hat{i}+b\hat{j}$

$\hat{n}$ is perpendicular to the vector $(\hat{i}+\hat{j})$

Let us take $(\hat{i}+\hat{j}) = \vec{B}$

The dot product of the two vectors are zero.

Therefore,

$a\hat{i}+b\hat{j}\cdot (\hat{i}+\hat{j})=0$

$a+b=0$

$a=-b$...(I)

We know that,

$\hat{n}=1$

Therefore, |n|=$\sqrt{a^2+b^2}=1$

$a^2+b^2=1$....(II)

Put the value of a in equation (II)

$(-b^2)+b^2=1$

$2b^2=1$

$b=\sqrt{\dfrac{1}{2}}$

Now, put the value of b in equation (I)

$a =-\sqrt{\dfrac{1}{2}}$

Hence, The value of a and b are $-\sqrt{\dfrac{1}{2}}, \sqrt{\dfrac{1}{2}}$

Answer 2

Hope you will understand it.....