Question

please solve these two maths questions....no inaccurate answers....✌​

Answer 1

Answer:

Step-by-step explanation:

7. the normal is equally inclined to coordinate axis

so, cosα=cosβ=cosψ

again, cos²α+cos²β+cos²ψ=1    ►3cos²α = 1  ►cosα = $\frac{1}{\sqrt{3} }$    =cosβ =cosψ

now, an unit vector parallel to the normal = $\frac{1}{\sqrt{3} }$  i + $\frac{1}{\sqrt{3} }$ j+ $\frac{1}{\sqrt{3} }$ k

Now, ($\frac{1}{\sqrt{3} }$  i + $\frac{1}{\sqrt{3} }$ j + $\frac{1}{\sqrt{3} }$ k)(x i + y j+ z k)  = 3$\sqrt{3}$

► x+y+z = 9

8. l+m+n=0  

l²+m²+n²=0  

here, sum of square of numbers is 0 , so, all the numbers will be 0 seperately

l=0, m= 0, n=0 (as square of a number is either positive or zero)

So, there is no answer

actually, the question has an error, l²+m²+n² =0 which can not be.

Answer 2

Step-by-step explanation:

Answer:

Step-by-step explanation:

7. the normal is equally inclined to coordinate axis

so, cosα=cosβ=cosψ

again, cos²α+cos²β+cos²ψ=1 ►3cos²α = 1 ►cosα = \frac{1}{\sqrt{3} }

3

1

=cosβ =cosψ

now, an unit vector parallel to the normal = \frac{1}{\sqrt{3} }

3

1

i + \frac{1}{\sqrt{3} }

3

1

j+ \frac{1}{\sqrt{3} }

3

1

k

Now, (\frac{1}{\sqrt{3} }

3

1

i + \frac{1}{\sqrt{3} }

3

1

j + \frac{1}{\sqrt{3} }

3

1

k)(x i + y j+ z k) = 3\sqrt{3}

3

► x+y+z = 9

8. l+m+n=0

l²+m²+n²=0

here, sum of square of numbers is 0 , so, all the numbers will be 0 seperately

l=0, m= 0, n=0 (as square of a number is either positive or zero)

So, there is no answer

actually, the question has an error, l²+m²+n² =0 which can not be.