In a parallelogram ABCD ,|A B|=a,|AD|=b and |AC|=c, the value of DB ⋅ AB is

Question

In a parallelogram ABCD ,|A B|=a,|AD|=b and |AC|=c, the value of DB ⋅ AB is (A) (3 a2+b2-c2/2) (B) (a2+3 b2-c2/2) (C) (a2-b2+3 c2/2) (D) (a2+3 b2+c2/2)

Tardigrade Admin · Accepted Answer

Let AB= veca, A D= vecb and AC= vecc when a , b and vecc are non-collinear coplanar vectors. DB = AB - AC = a - b Now, vecD B ⋅ AB=( veca- vecb) ⋅( veca)= veca ⋅ veca- vecb ⋅ veca a2-a b cos θ =a2-(c2-a2-b2/2)=(3 a2+b2-c2/2) [∵ In Δ ABC, cos (π-θ)=(a2+b2-c2/2 a b)] <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/9257ef7c262f6ccf145c59247b765d40-.png" />

Q. In a parallelogram $A BC D, ∣ A B ∣ = a, ∣ A D ∣ = b$ and $∣ A C ∣ = c$ , the value of $D B \cdot A B$ is

$\frac{3 a ^{2} + b ^{2} - c ^{2}}{2}$

$\frac{a ^{2} + 3 b ^{2} - c ^{2}}{2}$

$\frac{a ^{2} - b ^{2} + 3 c ^{2}}{2}$

$\frac{a ^{2} + 3 b ^{2} + c ^{2}}{2}$

Q. In a parallelogram ABCD,∣AB∣=a,∣AD∣=b and ∣AC∣=c, the value of DB⋅AB is

23a2+b2−c2​

2a2+3b2−c2​

2a2−b2+3c2​

2a2+3b2+c2​

Let AB=a,AD=b and AC=c when a,b and c are non-collinear coplanar vectors. DB=AB−AC=a−b Now, DB⋅AB=(a−b)⋅(a)=a⋅a−b⋅a a2−abcosθ=a2−2c2−a2−b2​=23a2+b2−c2​ [∵ In ΔABC,cos(π−θ)=2aba2+b2−c2​]

Q. In a parallelogram $A BC D, ∣ A B ∣ = a, ∣ A D ∣ = b$ and $∣ A C ∣ = c$ , the value of $D B \cdot A B$ is

$\frac{3 a ^{2} + b ^{2} - c ^{2}}{2}$

$\frac{a ^{2} + 3 b ^{2} - c ^{2}}{2}$

$\frac{a ^{2} - b ^{2} + 3 c ^{2}}{2}$

$\frac{a ^{2} + 3 b ^{2} + c ^{2}}{2}$