Question 25 of 68intermediate🔧 ApplyLong Answer4 marks

Show that each of the given three vectors is a unit vector: 17(2i^+3j^+6k^)\frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k}), 17(3i^−6j^+2k^)\frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k}), 17(6i^+2j^−3k^)\frac{1}{7}(6\hat{i} + 2\hat{j} - 3\hat{k}). Also, show that they are mutually perpendicular to each other.

Correct Answer

Let the given vectors be

a⃗=17(2i^+3j^+6k^)\vec{a} = \frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k}) b⃗=17(3i^−6j^+2k^)\vec{b} = \frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k}) c⃗=17(6i^+2j^−3k^)\vec{c} = \frac{1}{7}(6\hat{i} + 2\hat{j} - 3\hat{k})

Showing each is a unit vector:

The magnitude of a⃗\vec{a} is

∣a⃗∣=1722+32+62=174+9+36=1749=17×7=1|\vec{a}| = \frac{1}{7}\sqrt{2^2 + 3^2 + 6^2} = \frac{1}{7}\sqrt{4+9+36} = \frac{1}{7}\sqrt{49} = \frac{1}{7}\times 7 = 1

Similarly, magnitude of b⃗\vec{b} is

∣b⃗∣=1732+(−6)2+22=179+36+4=1749=1|\vec{b}| = \frac{1}{7}\sqrt{3^2 + (-6)^2 + 2^2} = \frac{1}{7}\sqrt{9+36+4} = \frac{1}{7}\sqrt{49} = 1

Magnitude of c⃗\vec{c} is

∣c⃗∣=1762+22+(−3)2=1736+4+9=1749=1|\vec{c}| = \frac{1}{7}\sqrt{6^2 + 2^2 + (-3)^2} = \frac{1}{7}\sqrt{36+4+9} = \frac{1}{7}\sqrt{49} = 1

Thus, each vector has magnitude 1, so they are unit vectors.

Showing mutual perpendicularity:

Two vectors are perpendicular if their dot product is zero.

a⃗⋅b⃗=17⋅17[(2)(3)+(3)(−6)+(6)(2)]=149(6−18+12)=0\vec{a}\cdot\vec{b} = \frac{1}{7}\cdot\frac{1}{7}\left[(2)(3)+(3)(-6)+(6)(2)\right] = \frac{1}{49}(6-18+12)=0 b⃗⋅c⃗=17⋅17[(3)(6)+(−6)(2)+(2)(−3)]=149(18−12−6)=0\vec{b}\cdot\vec{c} = \frac{1}{7}\cdot\frac{1}{7}\left[(3)(6)+(-6)(2)+(2)(-3)\right] = \frac{1}{49}(18-12-6)=0 c⃗⋅a⃗=17⋅17[(6)(2)+(2)(3)+(−3)(6)]=149(12+6−18)=0\vec{c}\cdot\vec{a} = \frac{1}{7}\cdot\frac{1}{7}\left[(6)(2)+(2)(3)+(-3)(6)\right] = \frac{1}{49}(12+6-18)=0

Since all pairwise dot products are zero, the vectors are mutually perpendicular.

Hence, each given vector is a unit vector and they are mutually perpendicular to each other.

Exercise: EXERCISE 10.3 | Q: 5 | (Page 24)
For More Understanding

Explanation

The solution uses the definition of magnitude of a vector to verify that each vector has length 1, confirming they are unit vectors. Then it uses the dot product criterion for perpendicularity: two vectors are perpendicular if and only if their dot product equals zero.

By computing the dot products of each pair and showing they are zero, the mutual perpendicularity is established. The calculations rely on the standard formulas for magnitude and dot product, which are consistent with the examples provided in the context (e.g., Example 7 and Example 8).

Solution Steps

  1. Step 1: Compute the magnitude of each vector using the formula ∣v⃗∣=x2+y2+z2|\vec{v}| = \sqrt{x^2 + y^2 + z^2} and show each equals 11.

  2. Step 2: Compute the dot product of each pair of vectors using the formula a⃗⋅b⃗=axbx+ayby+azbz\vec{a}\cdot\vec{b} = a_x b_x + a_y b_y + a_z b_z.

  3. Step 3: Show that each dot product equals zero, proving mutual perpendicularity.

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