Single correct+4 / -1Answer matches NTA’s final keyLet a⃗=4i^−j^+3k^\vec a = 4\hat i - \hat j + 3\hat ka=4i^−j^+3k^, b⃗=10i^+2j^−k^\vec b = 10\hat i + 2\hat j - \hat kb=10i^+2j^−k^ and a vector c⃗\vec cc be such that 2(a⃗×b⃗)+3(b⃗×c⃗)=0⃗2(\vec a \times \vec b) + 3(\vec b \times \vec c) = \vec 02(a×b)+3(b×c)=0. If a⃗⋅c⃗=15\vec a \cdot \vec c = 15a⋅c=15, then c⃗⋅(i^+j^−3k^)\vec c \cdot (\hat i + \hat j - 3\hat k)c⋅(i^+j^−3k^) is equal to: Choose oneA−6-6−6B−5-5−5C−4-4−4D−3-3−3Check answerShow solutionSolutionSolution2(a⃗×b⃗)=3(c⃗×b⃗)2(\vec a \times \vec b) = 3(\vec c \times \vec b)2(a×b)=3(c×b) gives (3c⃗−2a⃗)×b⃗=0⃗(3\vec c - 2\vec a) \times \vec b = \vec 0(3c−2a)×b=0, so 3c⃗−2a⃗=λb⃗3\vec c - 2\vec a = \lambda \vec b3c−2a=λb: c⃗=2a⃗+λb⃗3.\vec c = \frac{2\vec a + \lambda \vec b}{3}.c=32a+λb. Dot with a⃗\vec aa: ∣a⃗∣2=26|\vec a|^2 = 26∣a∣2=26 and a⃗⋅b⃗=40−2−3=35\vec a \cdot \vec b = 40 - 2 - 3 = 35a⋅b=40−2−3=35, so 15=2(26)+35λ3⇒35λ=−7⇒λ=−15.15 = \frac{2(26) + 35\lambda}{3} \Rightarrow 35\lambda = -7 \Rightarrow \lambda = -\tfrac15.15=32(26)+35λ⇒35λ=−7⇒λ=−51. Now with d⃗=i^+j^−3k^\vec d = \hat i + \hat j - 3\hat kd=i^+j^−3k^: a⃗⋅d⃗=4−1−9=−6\vec a \cdot \vec d = 4 - 1 - 9 = -6a⋅d=4−1−9=−6 and b⃗⋅d⃗=10+2+3=15\vec b \cdot \vec d = 10 + 2 + 3 = 15b⋅d=10+2+3=15: c⃗⋅d⃗=2(−6)+(−15)(15)3=−12−33=−5.\vec c \cdot \vec d = \frac{2(-6) + (-\tfrac15)(15)}{3} = \frac{-12 - 3}{3} = -5.c⋅d=32(−6)+(−51)(15)=3−12−3=−5. Answer: −5-5−5. Answer checked against the official answer key. The solution is our own working.