Explanation
Step 1: c ki direction nikalna
Hame diya gaya hai:
Isse expand karne par:
Chunki c×c=0 hota hai, iska matlab:
Jab do vectors ka cross product zero hota hai, toh woh aapas mein parallel hote hain. Isliye:
Jahan k ek scalar constant hai.
Step 2: a+b ki value nikalna
Toh, c=k((λ+3)i^+k^).
Step 3: λ aur k ki values nikalna
Di gayi conditions ka use karte hain:
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b⋅c=−20
(3i^−j^+2k^)⋅k((λ+3)i^+k^)=−20
k[3(λ+3)+2(1)]=−20⟹k(3λ+11)=−20—(i)
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a⋅c=−17
(λi^+j^−k^)⋅k((λ+3)i^+k^)=−17
k[λ(λ+3)−1]=−17⟹k(λ2+3λ−1)=−17—(ii)
Dono equations ko divide karne par:
3λ+11λ2+3λ−1=−20−17=2017
Is quadratic ko solve karne par λ=3 (kyunki λ∈Z).
Ab k ki value nikalte hain:
k(3(3)+11)=−20⟹20k=−20⟹k=−1
Toh, c=−1((3+3)i^+k^)=−6i^−k^.
Step 4: Final Value Calculate karna
Hame ∣c×(λi^+j^+k^)∣2 nikalna hai, jahan λ=3:
c×v=i^−63amp;j^amp;0amp;1amp;k^amp;−1amp;1=i^(0+1)−j^(−6+3)+k^(−6−0)=i^+3j^−6k^
Iska magnitude ka square:
∣c×v∣2=12+32+(−6)2=1+9+36=46
