If abc are three non-coplanar unit vectors then -abc- is equal to

The correct option is B ± 1If abc are three non coplanar unit vectors then [abc]=± 1 nbsp; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Scalar Triple Product

If abc are th...

Question

If $\to a \to b \to c$ are three non-coplanar unit vectors then $[\to a \to b \to c]$ is equal to

\pm 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\pm 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\pm 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

\vec{a,} \vec{b,} \vec{c}

[\vec{a} \vec{b} \vec{c}],

| \to a | = 1, | \to b | = 2

| \to c | = 3

\to a, \to b, \to c

| \to d | = 4,

∣ ∣ ∣ ∣ ∣ ∣ \frac{[\to d \to b \to c] \to a + [\to d \to c \to a] \to b + [\to d \to a \to b] \to c}{[\to a \to b \to c]} ∣ ∣ ∣ ∣ ∣ ∣

If $\to a, \to b$ and $\to c$ are three non-coplanar vectors, then $(\to a + \to b + \to c) \cdot [(\to a + \to b) \times (\to a + \to c)]$ equals

∣ ∣ ∣ ∣ ∣ \begin{matrix} a & a^{2} & 1 + a^{3} b & b^{2} & 1 + b^{3} c & c^{2} & 1 + c^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

(1, a, a^{2}), (1, b, b^{2})

(1, c, c^{2})

a b c

\to a, \to b, \to c

\to p, \to q, \to r

\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]}, \to r = \frac{\to a \times \to b}{[\to a \to b \to c]},

(\to a + \to b + \to c), (\to p + \to q + \to r)

Join BYJU'S Learning Program