Question

If \(\bar{a}\times\bar{b}=7\hat{i}+9\hat{j}+10\hat{k}\) and \(\bar{a}\cdot\bar{b}=-20\), then \(|\bar{a}|^{2}|\bar{b}|^{2}=\)

• A. 530
• B. 580
• C. 400
• D. 630
• E. 560

Answer 1

The Correct Option is D

We are given:
\( \mathbf{a} \times \mathbf{b} = 7\hat{i} + 9\hat{j} + 10\hat{k} \) - \( \mathbf{a} \cdot \mathbf{b} = -20 \) We need to calculate \( |\mathbf{a}|^2 |\mathbf{b}|^2 \).
Step 1: Use the identity for the magnitude of the cross product The magnitude of the cross product is given by: \[ |\mathbf{a} \times \mathbf{b}| = \sqrt{7^2 + 9^2 + 10^2} = \sqrt{49 + 81 + 100} = \sqrt{230} \] So, \[ |\mathbf{a} \times \mathbf{b}| = \sqrt{230} \]
Step 2: Use the identity for the magnitude of the cross product We know that: \[ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin \theta \] Where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \). Thus, we have: \[ \sqrt{230} = |\mathbf{a}| |\mathbf{b}| \sin \theta \] ### Step 3: Use the dot product identity We are also given: \[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta = -20 \]
Step 4: Solve for \( |\mathbf{a}|^2 |\mathbf{b}|^2 \) Now we have the two equations: 1. \( |\mathbf{a}| |\mathbf{b}| \sin \theta = \sqrt{230} \) 2. \( |\mathbf{a}| |\mathbf{b}| \cos \theta = -20 \) Squaring both equations: \[ (|\mathbf{a}| |\mathbf{b}| \sin \theta)^2 = 230 \] \[ (|\mathbf{a}| |\mathbf{b}| \cos \theta)^2 = 400 \] Adding these equations: \[ (|\mathbf{a}| |\mathbf{b}| \sin \theta)^2 + (|\mathbf{a}| |\mathbf{b}| \cos \theta)^2 = 230 + 400 \] \[ |\mathbf{a}|^2 |\mathbf{b}|^2 (\sin^2 \theta + \cos^2 \theta) = 630 \] Since \( \sin^2 \theta + \cos^2 \theta = 1 \), we have: \[ |\mathbf{a}|^2 |\mathbf{b}|^2 = 630 \]

The correct option is (D) : \(630\)

Answer 2

We are given \(\bar{a}\times\bar{b} = 7\hat{i} + 9\hat{j} + 10\hat{k}\) and \(\bar{a}\cdot\bar{b} = -20\).

We know that \(|\bar{a}\times\bar{b}| = |\bar{a}||\bar{b}|\sin\theta\) and \(\bar{a}\cdot\bar{b} = |\bar{a}||\bar{b}|\cos\theta\), where \(\theta\) is the angle between \(\bar{a}\) and \(\bar{b}\).

We can find \(|\bar{a}\times\bar{b}|\) using the components:

\(|\bar{a}\times\bar{b}| = \sqrt{7^2 + 9^2 + 10^2} = \sqrt{49 + 81 + 100} = \sqrt{230}\)

So, \(|\bar{a}||\bar{b}|\sin\theta = \sqrt{230}\) and \(|\bar{a}||\bar{b}|\cos\theta = -20\).

We want to find \(|\bar{a}|^2|\bar{b}|^2 = (|\bar{a}||\bar{b}|)^2\). We can use the identity \(\sin^2\theta + \cos^2\theta = 1\):

\((|\bar{a}||\bar{b}|\sin\theta)^2 + (|\bar{a}||\bar{b}|\cos\theta)^2 = (|\bar{a}||\bar{b}|)^2(\sin^2\theta + \cos^2\theta) = (|\bar{a}||\bar{b}|)^2\)

Therefore,

\((|\bar{a}||\bar{b}|)^2 = (\sqrt{230})^2 + (-20)^2 = 230 + 400 = 630\)

So, \(|\bar{a}|^2|\bar{b}|^2 = 630\).