Let vector r lie in the plane of p = i + j and q = j + k, and be such that r is perpendicular to p and satisfies r · q = 2. Then, the value of the dot product r · (2i + 3j +
k) is:

We are given:

• Vector p = i + j
• Vector q = j + k
• Vector r lies in the plane of p and q, so r = αp + βq = α(i + j) + β(j + k)
• Given conditions:

1.     r ⊥ p ⇒ r · p = 0 2.     r · q = 2

• Find: r · (2i + 3j + k)

Express r in terms of α and β r = α(i + j) + β(j + k) ⇒ r = αi + (α + β)j + βk Apply condition 1: r · p = 0 p = i + j So, r · p = (αi + (α + β)j + βk) · (i + j) = α(1) + (α + β)(1) + β(0) = α + α + β = 2α + β = 0 ⇒ 2α + β = 0 … (1) Apply condition 2: r · q = 2 q = j + k So, r · q = (αi + (α + β)j + βk) · (j + k) = (α)(0) + (α + β)(1) + β(1) = (α + β) + β = α + 2β = 2 ⇒ α + 2β = 2 … (2) Solve the system of equations From (1): β = –2α Substitute into (2): α + 2(–2α) = 2 α – 4α = 2 –3α = 2 ⇒ α = –2/3 Then β = –2α = –2(–2/3) = 4/3 Find r r = αi + (α + β)j + βk = (–2/3)i + (–2/3 + 4/3)j + (4/3)k = (–2/3)i + (2/3)j + (4/3)k Compute r · (2i + 3j + k) r · (2i + 3j + k) = (–2/3)(2) + (2/3)(3) + (4/3)(1) = –4/3 + 6/3 + 4/3 = 6/3 = 2