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JEE MATHEMATICS MCQ: VECTOR TRIPLE PRODUCT

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Let a=i^+j^+2k^a = \hat i + \hat j + 2\hat k and b=−i^+2j^+3k^b = - \hat i + 2\hat j + 3\hat k. Then the vector product (a+b)×[(a×{(a−b)×b)}×b] (a + b) \times [(a \times \{ (a - b) \times b)\} \times b]\, is equal to
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Correct option: B
7 (34i^−5j^+3k^)7\,(34\hat i - 5\hat j + 3\hat k)
Option B (7(34i^−5j^+3k^)7(34\hat i - 5\hat j + 3\hat k)) is correct because it is the result of simplifying the vector triple product and substituting the given vectors.

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Let u^{\mathbf{\hat u}} and v^{\mathbf{\hat v}} are unit vectors such that u^⋅v^=0\mathbf{\hat u}\cdot\mathbf{\hat v} = 0 If r^{\mathbf{\hat r}} is any vector coplanar with u^{\mathbf{\hat u}} and v^{\mathbf{\hat v}}, then the magnitude of the vector r⃗×(u^×v^)\mathbf{\vec r} \times \left( {\mathbf{\hat u} \times \mathbf{\hat v}} \right) is Let λ∈R,a⃗=λi^+2j^−3k^,b⃗=i^−λj^+2k^\lambda \in \mathrm{R}, \vec{a}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}, \vec{b}=\hat{i}-\lambda \hat{j}+2 \hat{k}
If ((a⃗+b⃗)×(a⃗×b⃗))×(a⃗−b⃗)=8i^−40j^−24k^((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})=8 \hat{i}-40 \hat{j}-24 \hat{k}
then ∣λ(a⃗+b⃗)×(a⃗−b⃗)∣2|\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2 is equal to
Let a⃗=−i^−j^+k^\vec{a}=-\hat{i}-\hat{j}+\hat{k} such that a⃗⋅b⃗=1\vec{a} \cdot \vec{b}=1 and a⃗×b⃗=i^−j^\vec{a} \times \vec{b}=\hat{i}-\hat{j} then a⃗−6b⃗\vec{a}-6 \vec{b} equals Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} be any three vectors. Then, vectors u⃗=a⃗×(b⃗×c⃗),v⃗=b⃗×(c⃗×a⃗)   and   w⃗=c⃗×(a⃗×b⃗)\vec u = \vec a \times (\vec b \times \vec c),\vec v = \vec b \times (\vec c \times \vec a)\,\,\,{\text{and}}\,\,\,\vec w = \vec c \times (\vec a \times \vec b) are such that they are If aa and bb are perpendicular, then a ×(a ×(a ×(a ×b)))a\, \times (a\, \times (a\, \times (a\, \times b))) is equal to For any vector a⃗\vec{a}, the value of i^×(a⃗×i⃗)+j^×(a⃗×j^)+k^×(a⃗×k⃗)\hat{i} \times(\vec{a} \times \vec{i})+\hat{j} \times(\vec{a} \times \hat{j})+\hat{k} \times(\vec{a} \times \vec{k}), is
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