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If θ\theta is angle between a⃗\vec{a} & b⃗\vec{b}, then for vectors a⃗,b⃗\vec{a}, \vec{b} and c⃗\vec{c},

Match the LIST-I with LIST-II

LIST-ILIST-II
A.a⃗×b⃗=a⃗×c⃗,a⃗≠0,b⃗≠c⃗\vec{a} \times \vec{b} = \vec{a} \times \vec{c}, \vec{a} \neq 0, \vec{b} \neq \vec{c}I.θ=π4\theta = \dfrac{\pi}{4}
B.a⃗×b⃗=0\vec{a} \times \vec{b} = 0II.a⃗∥(b⃗−c⃗)\vec{a} \parallel (\vec{b} - \vec{c})
C.∣a⃗.b⃗∣=∣a⃗×b⃗∣\vert \vec{a}.\vec{b} \vert = \vert \vec{a} \times \vec{b} \vertIII.b⃗=c⃗\vec{b} = \vec{c}
D.a⃗.b⃗=a⃗.c⃗\vec{a}.\vec{b} = \vec{a}.\vec{c} and a⃗×b⃗=a⃗×c⃗\vec{a} \times \vec{b} = \vec{a} \times \vec{c}, a⃗≠0\vec{a} \neq 0IV.a⃗=0,b⃗=0\vec{a} = 0, \vec{b} = 0 or a⃗∥b⃗\vec{a} \parallel \vec{b}

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 4

From a⃗×(b⃗−c⃗)=0\vec a\times(\vec b-\vec c)=0 with b⃗≠c⃗\vec b\neq\vec c, a⃗\vec a is parallel to b⃗−c⃗\vec b-\vec c (II). A zero cross product means a zero vector or parallel vectors (IV). ∣a⃗⋅b⃗∣=∣a⃗×b⃗∣|\vec a\cdot\vec b|=|\vec a\times\vec b| gives tan⁡θ=1\tan\theta=1, so θ=π4\theta=\frac{\pi}{4} (I). Both products equal forces b⃗=c⃗\vec b=\vec c (III).

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