Question

consider the value of two vectors L=1i+2j+3k and l=4i+5j+6k.Find the value of the scalar A such that the vector L-Al is perpendicular to the L​

Answer 1

Given:

L=1i+2j+3k

I=4i+5j+6k

To Find :

Find a scalar A such that L-AI is perpendicular to L.

(L-AI).L=0

Explanation:

AI=A(4i+5j+6k)

L-AI= 1i+2j+3k - A(4i+5j+6k)

      =(1-4A)i + (2-5A)j + (3-6A)k

(L-AI).L={(1-4A)i + (2-5A)j + (3-6A)k}.(1i+2j+3k)

          =(1-4A)x1 + (2-5A)x2 + (3-6A)x3     (i.i=1.j.j=1,k.k=1)

          =1-4A + 4 -10A +9- 18A

          =14-32A

(L-AI).L =0

∴14-32A = 0

 A= 7/16

Hence the value of the scalar A is 7/16.

     

Answer 2

Answer:

Value of scalar is.

Explanation:

Given:

$L= i+2j+3k\\I= 4i+5j+6k$

A is a scalar.

To find:

The value of scalar is.

Calculation:

As we know that if two vectors are perpendicular to each other that their dot product is equal to zero.

$L.(L-AI)=0\\(1i+2j+3k).(L-A(4i+5j+6k))=0\\(1i+2j+3k).[(1i+2j+3k)-(4iA+5jA+6kA)]=0\\(1i+2j+3k).(i(1-4A)+j(2-5A)+k(3-6A))=0\\(1-4A)+2(2-5A)+3(3-6A)=0\\1-4A+4-10A+9-18A=0\\14-32A=0\\32A=14\\A=\frac{14}{32} \\A=\frac{7}{16}$

Hence, value of Scalar is $A=\frac{7}{16}$