(a+b)×c−c×(−2a+3b)=0(a+b)×c+(−2a+3b)×c=0⇒(a+b)−2a+3b)×c=0⇒c=λ(4b−a)⇒=λ(44i^−4j^+4k^−4i^+j^−k^)=λ(40i^−3j^+3k^) Now $\begin{aligned}
& (8 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}+33 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot \lambda(40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=1670 \
& \Rightarrow(41 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \cdot(40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \times \lambda=1670) \
& \Rightarrow(1640+15+15) \lambda=1670 \Rightarrow \lambda=1 \
& \text { so } \overrightarrow{\mathrm{c}}=40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \
& \Rightarrow|\overrightarrow{\mathrm{c}}|^2=1600+9+9=1618
\end{aligned}$
Let a=4 i- j+ k, b=11 i- j+ k and be a vector such that ( a+ b) × = ×(-2 a+3 b). If (2 a+3 b) · =1670, then | |^2 is equal to :
JEE Main 2024 — Mathematics Vectors & 3D Geometry
2024mcqmedium
Let a=4i^−j^+k^,b=11i^−j^+k^ and c be a vector such that (a+b)×c=c×(−2a+3b). If (2a+3b)⋅c=1670, then ∣c∣2 is equal to :
Official previous-year question
Held on 8 Apr 2024 · Verified 6 Jul 2026.
Options
- A
1609
- B
1618
- C
1600
- D
1627
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