Question

The volume of the tetrahedron whose co-terminus edges are $\bar{a}, \bar{b}, \bar{c}$ is 12 cubic units. If the scalar projection of $\bar{a}$ on $\bar{b} \times \bar{c}$ is 4 , then $|\bar{b} \times \bar{c}| =$

• A. 18
• B. $\frac{1}{18}$
• C. 16
• D. $\frac{1}{16}$

Hint:

Volume formula: $V = \frac{1}{6}|\vec{a}\cdot(\vec{b}\times\vec{c})|$

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The volume of a tetrahedron formed by three co-terminus vectors is related to their scalar triple product. The scalar projection of one vector onto another provides a relation between their dot product and magnitude. We can link these two concepts because the scalar triple product $[\bar{a} \quad \bar{b} \quad \bar{c}]$ is exactly the dot product of $\bar{a}$ and $(\bar{b} \times \bar{c})$. Step 2: Key Formula or Approach:
1. Volume of tetrahedron: $V = \frac{1}{6} |[\bar{a} \quad \bar{b} \quad \bar{c}]| = \frac{1}{6} |\bar{a} \cdot (\bar{b} \times \bar{c})|$. 2. Scalar projection of vector $\bar{u}$ on vector $\bar{v}$ is given by $\frac{|\bar{u} \cdot \bar{v}|}{|\bar{v}|}$. Here, let $\bar{v} = \bar{b} \times \bar{c}$. Step 3: Detailed Explanation:
Given the volume of the tetrahedron is 12: \[ \frac{1}{6} |[\bar{a} \quad \bar{b} \quad \bar{c}]| = 12 \] \[ |[\bar{a} \quad \bar{b} \quad \bar{c}]| = 12 \times 6 = 72 \] We know that the scalar triple product can be written as a dot product with a cross product: \[ |[\bar{a} \quad \bar{b} \quad \bar{c}]| = |\bar{a} \cdot (\bar{b} \times \bar{c})| = 72 \] The scalar projection of $\bar{a}$ on the vector $(\bar{b} \times \bar{c})$ is given as 4. The formula for the scalar projection of $\bar{a}$ on $\bar{n}$ (where $\bar{n} = \bar{b} \times \bar{c}$) is: \[ \text{Projection} = \frac{|\bar{a} \cdot \bar{n}|}{|\bar{n}|} \] Substitute $\bar{n} = \bar{b} \times \bar{c}$ and the given projection value: \[ \frac{|\bar{a} \cdot (\bar{b} \times \bar{c})|}{|\bar{b} \times \bar{c}|} = 4 \] We already found the numerator to be 72: \[ \frac{72}{|\bar{b} \times \bar{c}|} = 4 \] Solve for the magnitude $|\bar{b} \times \bar{c}|$: \[ |\bar{b} \times \bar{c}| = \frac{72}{4} \] \[ |\bar{b} \times \bar{c}| = 18 \] Step 4: Final Answer:
The value of $|\bar{b} \times \bar{c}|$ is 18.