Let A, B, C and D are 4 points in space satisfying the equation AB→⋅CD→=kAD→^2+BC→^2-AC→^2-BD→^2, then the value of k is equal to

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MARKS Content Team · Accepted Answer

Correct answer: (A) 1/2. Let position vectors of A,B,C,D are a→,b→,c→,d→ respectively AB→⋅CD→=b→-a→⋅d→-c→ =b→⋅d→-b→⋅c→-a→⋅d→+a→⋅c→ … 1 AB→^2+BC→^2-AC→^2-BD→^2 =d→^2+a→^2-2a→⋅d→+c→^2+b→^2-2b→⋅c→-a→^2-c→^2+2a→⋅c→-b→^2-d→^2+2b→⋅d→ =2b→⋅d→-b→⋅c→-a→⋅d→+a→⋅c→ … 2 From 1 & 2, k=0.5

Let $A, B, C$ and $D$ are $4$ points in space satisfying the equation $\vec{A B} \cdot \vec{C D} = k ({|\vec{A D}|}^{2} + {|\vec{B C}|}^{2} - {|\vec{A C}|}^{2} - {|\vec{B D}|}^{2})$ , then the value of $k$ is equal to

Solution

More Vector Algebra Questions

Let A, B, C and D are 4 points in space satisfying the equation AB→⋅CD→=kAD→2+BC→2-AC→2-BD→2, then the value of k is equal to

Solution

More Vector Algebra Questions

Let $A, B, C$ and $D$ are $4$ points in space satisfying the equation $\vec{A B} \cdot \vec{C D} = k ({|\vec{A D}|}^{2} + {|\vec{B C}|}^{2} - {|\vec{A C}|}^{2} - {|\vec{B D}|}^{2})$ , then the value of $k$ is equal to