Question

please give me answer of this question As soon as possible.​

Answer 1

the answer is 904

AWM KAR98K VECTOR

Answer 2

Step-by-step explanation:

STEP 1.The position vector of point (0,1,−2) is a→=j^−2k^

The normal vector N→ perpendicular to the plane is N→=i^+j^+k^

We know the vector equation of the plane is given by (r→−a→).N→=0

STEP 2.to the plane is N→=i^+j^−k^

We know the vector equation of the plane is given by (r→−a→).N→=0

Step 2:

On substituting for a→ and N→ we get,

[r→ − (j^−2k^)].(i^+j^+k^)=0

But we know r→=xi^+yj^+zk^

Therefore [(xi^+yj^+zk^)−(j^−2k^)].(i^+j^+k^)=0

STEP 3. On simplifying we get,

[x^+(y-1)j^+(z+2)k^)].(i^+j^+k^)=0

We know that i^.i^=j^.j^=k^.k^=1

⇒x + (y−1) + (z+2)=0

x+y+z+1=0

x+y+z= -1

Therefore x+y+z=-1 is the required Cartesian equation of the plane.