Question

Let \(\overrightarrow a=\overrightarrow i+2\overrightarrow j+3 \overrightarrow k\) and \(\overrightarrow b= \overrightarrow i+ \overrightarrow j - \overrightarrow k\) . If \(\overrightarrow c\) is a vector such that \(\overrightarrow a. \overrightarrow c=11,\overrightarrow b.(\overrightarrow a.\overrightarrow c)=27\) and \(\overrightarrow b. \overrightarrow c=-\sqrt{3}|b|,\)  then \(|\overrightarrow a \times \overrightarrow c|^2\) is equal to _______ .

Answer 1

Correct Answer: 285

We are given:

\( \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{b} = \hat{i} + \hat{j} - \hat{k}. \) 

Step 1: Properties of the vectors

1. The dot product: \( \vec{a} \cdot \vec{b} = 0 \) This implies that \( \vec{a} \) and \( \vec{b} \) are perpendicular.
2. The cross product: \( \vec{b} \times (\vec{a} \times \vec{c}) = -3\vec{a}. \)

Step 2: Relationship involving angle \( \theta \)

Let \( \theta \) be the angle between \( \vec{b} \) and \( \vec{a} \times \vec{c} \). The magnitude of their product is:

\( |\vec{b} \cdot (\vec{a} \times \vec{c})| \sin \theta = 3\sqrt{14}. \)

The magnitude of the dot product is:

\( |\vec{b} \cdot (\vec{a} \times \vec{c})| \cos \theta = 27. \)

Step 3: Solve for \( \sin \theta \)

Divide the equations to find \( \sin \theta \):

\( \sin \theta = \frac{\sqrt{14}}{\sqrt{95}}. \)

Step 4: Find \( |\vec{b} \times (\vec{a} \times \vec{c})| \)

From the given relationships:

\( |\vec{b} \times (\vec{a} \times \vec{c})| = 3\sqrt{95}. \)

Step 5: Find \( |\vec{a} \times \vec{c}| \)

From the magnitude relationship:

\( |\vec{a} \times \vec{c}| = \sqrt{3 \times \sqrt{95}}. \)

\( |\vec{a} \times \vec{c}|^2 = 3 \times 95 = 285. \)

Final Answer:

\( |\vec{a} \times \vec{c}| = \sqrt{285}. \)