If $\vec a\cdot\vec b=\vec b\cdot\vec c=\vec c\cdot\vec a=0$, then the value of $[\vec a,\vec b,\vec c]$ is (1) $|\vec a||\vec b||\vec c|$ (2) $\tfrac13|\vec a||\vec b||\vec c|$ (3) $1$ (4) $-1$

If vec a·vec b=vec b·vec c=vec c·vec a=0, then the value of [vec a,vec b,vec c] is (1) |vec a||vec b||vec c| (

Exercise 6.10 · Q3

\vec a\cdot\vec b=\vec b\cdot\vec c=\vec c\cdot\vec a=0

(1)

|\vec a||\vec b||\vec c|

(2)

\tfrac13|\vec a||\vec b||\vec c|

(3)

1

(4)

-1

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✓ Free question

When three vectors are pairwise perpendicular, the unique direction perpendicular to two of them (their cross product) must be parallel to the third — so the scalar triple product reduces to the product of the three magnitudes.

Step 1. Interpret the hypothesis. $\vec a\cdot\vec b=0,\ \vec b\cdot\vec c=0,\ \vec c\cdot\vec a=0$ means $\vec a,\vec b,\vec c$ are mutually (pairwise) perpendicular.

Step 2. Since $\vec b\perp\vec c$ , $\vec b\times\vec c$ has magnitude $|\vec b||\vec c|$ and points along the unique direction perpendicular to both $\vec b$ and $\vec c$ .

Step 3. Since $\vec a$ is ALSO perpendicular to both $\vec b$ and $\vec c$ , $\vec a$ must be parallel to that same unique direction, i.e. $\vec a\parallel(\vec b\times\vec c)$ .

Step 4. Compute $[\vec a,\vec b,\vec c]=\vec a\cdot(\vec b\times\vec c)$ . Since $\vec a$ is parallel to $\vec b\times\vec c$ : $\vec a\cdot(\vec b\times\vec c)=\pm|\vec a||\vec b\times\vec c|=\pm|\vec a||\vec b||\vec c|$ (taking the positive/standard orientation as intended by the option).

✓Final answer

[\vec a,\vec b,\vec c]=\boxed{|\vec a||\vec b||\vec c|}

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Q.If $\vec a\cdot\vec b=\vec b\cdot\vec c=\vec c\cdot\vec a=0$ , then the value of $[\vec a,\vec b,\vec c]$ is

Concept understanding — Scalar Triple Product and Coplanarity

Scalar Triple Product and Coplanarity

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Q.If $\vec a\cdot\vec b=\vec b\cdot\vec c=\vec c\cdot\vec a=0$ , then the value of $[\vec a,\vec b,\vec c]$ is

Unlock everything free for 14 days