CUET PG 2023 — Mathematics PYQ
CUET PG | Mathematics | 2023a=2i^+2j^+3k^,b=−i^+2j^+k^ and c=3i^+j^ are such that a+yb is perpendicular to c then determine the value of y?
Choose the correct answer:
- A.
3
- B.
0
- C.
4
- D.
8
(Correct Answer)
8
Explanation
Step 1: Find the vector a+yb
Given:
-
a=2i^+2j^+3k^
-
b=−i^+2j^+k^
Adding them with the scalar y:
a+yb=(2i^+2j^+3k^)+y(−i^+2j^+k^)
a+yb=(2−y)i^+(2+2y)j^+(3+y)k^
Step 2: Set the Dot Product with c to Zero
Since (a+yb)⊥c, we have:
(a+yb)⋅c=0
Given c=3i^+j^+0k^, we substitute the components:
[(2−y)i^+(2+2y)j^+(3+y)k^]⋅[3i^+1j^+0k^]=0
Multiplying corresponding components:
3(2−y)+1(2+2y)+0(3+y)=0
Step 3: Solve for y
Expand the equation:
6−3y+2+2y=0
Combine the like terms:
8−y=0
y=8
Answer: The value of y is 8.
Explanation
Step 1: Find the vector a+yb
Given:
-
a=2i^+2j^+3k^
-
b=−i^+2j^+k^
Adding them with the scalar y:
a+yb=(2i^+2j^+3k^)+y(−i^+2j^+k^)
a+yb=(2−y)i^+(2+2y)j^+(3+y)k^
Step 2: Set the Dot Product with c to Zero
Since (a+yb)⊥c, we have:
(a+yb)⋅c=0
Given c=3i^+j^+0k^, we substitute the components:
[(2−y)i^+(2+2y)j^+(3+y)k^]⋅[3i^+1j^+0k^]=0
Multiplying corresponding components:
3(2−y)+1(2+2y)+0(3+y)=0
Step 3: Solve for y
Expand the equation:
6−3y+2+2y=0
Combine the like terms:
8−y=0
y=8
Answer: The value of y is 8.

