In a ||gm ABCD;|AB|=a,|AD|=b and |AC|=c. Then DB .AB has the value

Question

In a ||gm ABCD;|AB|=a,|AD|=b and |AC|=c. Then DB .AB has the value (A) (3a2+b2-c2/2) (B) (a2+3b2-c2/2) (C) (a2-b2+3c2/2) (D) (a2+3b2+c2/2)

Tardigrade Admin · Accepted Answer

Since DB=DA+AB ∴ DB=DB-AB <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/e2da2ddb6c21cdb361329c678726b1d9-.jpeg" /> [Hence AB= veca, AD= vecb, AC= vecc] ∴ (DA)2+(DB)2+(AB)2-2 DB⋅AB ldots(1) Also in the || gm 2(a2+b2)=c2+DB2 ∴ DB2=2a2+2b2-c2 Putting in (1), b2=2a2+2b2-c2+a2-2 AB⋅DB ∴ AB⋅DB=(3a2+b2-c2/2)

Q. In a $∣∣ g m A BC D; ∣ A B ∣ = a, ∣ A D ∣ = b$ and $∣ A C ∣ = c .$ Then $D B . A B$ has the value

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

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

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

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

Q. In a ∣∣gmABCD;∣AB∣=a,∣AD∣=b and ∣AC∣=c. Then DB.AB has the value

23a2+b2−c2​

2a2+3b2−c2​

2a2−b2+3c2​

2a2+3b2+c2​

Since DB=DA+AB ∴DB=DB−AB [Hence AB=a,AD=b,AC=c] ∴(DA)2+(DB)2+(AB)2−2DB⋅AB…(1) Also in the ∣∣gm 2(a2+b2)=c2+DB2 ∴DB2=2a2+2b2−c2 Putting in (1), b2=2a2+2b2−c2+a2−2AB⋅DB ∴AB⋅DB=23a2+b2−c2​

Q. In a $∣∣ g m A BC D; ∣ A B ∣ = a, ∣ A D ∣ = b$ and $∣ A C ∣ = c .$ Then $D B . A B$ has the value

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

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

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

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