Question 38 of 68beginner🔧 ApplyNumerical2 marks

Find |a⃗ × b⃗|, if a⃗ = î - 7ĵ + 7k̂ and b⃗ = 3î - 2ĵ + 2k̂.

Correct Answer

722\sqrt{722}

Exercise: EXERCISE 10.4 | Q: 1 | (Page 31)
For More Understanding

Explanation

This question is solved using the determinant method for the cross product of two vectors, as illustrated in Example 22 of the provided text. First, the cross product a⃗×b⃗\vec{a} \times \vec{b} is calculated using the matrix determinant form. Then, the magnitude is found by taking the square root of the sum of the squares of the resulting vector's components.

Solution Steps

  1. Step 1: Compute the cross product a⃗×b⃗\vec{a} \times \vec{b} using the determinant method.

    a⃗×b⃗=∣i^j^k^1−773−22∣\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{vmatrix}

  2. Step 2: Expand the determinant to find the vector.

    =i^((−7)(2)−(7)(−2))−j^((1)(2)−(7)(3))+k^((1)(−2)−(−7)(3))= \hat{i}((-7)(2) - (7)(-2)) - \hat{j}((1)(2) - (7)(3)) + \hat{k}((1)(-2) - (-7)(3)) =i^(−14+14)−j^(2−21)+k^(−2+21)= \hat{i}(-14 + 14) - \hat{j}(2 - 21) + \hat{k}(-2 + 21) =0i^+19j^+19k^= 0\hat{i} + 19\hat{j} + 19\hat{k}

  3. Step 3: Calculate the magnitude of the resulting vector.

    ∣a⃗×b⃗∣=02+192+192|\vec{a} \times \vec{b}| = \sqrt{0^2 + 19^2 + 19^2} =0+361+361= \sqrt{0 + 361 + 361} =722= \sqrt{722}
PreviousNext