If a is a non-zero vector, then always Option 1: a.a = 0 Option 2: a.a > 0 Option 3: a.a < 0 Option 4: a.a = 1

Answer: a.a > 0. Solution: When taking the dot product of any vector a with itself: a.a =, a, ^2 where, a, is the magnitude of the vector. The dot product formula is: a.b =, a, b, When computing a.a, the angle between the vector and itself is = 0°: a.a =, a, a, (0°) =, a, ^2 x 1 =, a, ^2 Since a is a non-zero vector, its magnitude, a, > 0. Squaring any positive number gives a positive result. Therefore: a.a =, a, ^2 > 0 For a = 3i + 4j: Magnitude:, a, = √(3^2 + 4^2) = √(9 + 16) = 5 a.a = (3i + 4j).(3i + 4j) = 9 + 16 = 25 = 5^2 > 0 a.a = 0 only when a is a zero vector. a.a a.a > 0 is always true for any non-zero vector.

Algebra

Vector Algebra

Easy

core

If $\vec{a}$ is a non-zero vector, then always

$\vec{a}.\vec{a} = 0$

$\vec{a}.\vec{a} > 0$

$\vec{a}.\vec{a} < 0$

$\vec{a}.\vec{a} = 1$

✅ Correct Option: 2

When taking the dot product of any vector $\vec{a}$ with itself:

$\vec{a}.\vec{a} = |\vec{a}|^2$

where $|\vec{a}|$ is the magnitude of the vector.

The dot product formula is:

$\vec{a}.\vec{b} = |\vec{a}||\vec{b}|\cos\theta$

When computing $\vec{a}.\vec{a}$ , the angle between the vector and itself is $\theta = 0°$ :

$\vec{a}.\vec{a} = |\vec{a}||\vec{a}|\cos(0°)$

$= |\vec{a}|^2 \times 1$

$= |\vec{a}|^2$

Since $\vec{a}$ is a non-zero vector, its magnitude $|\vec{a}| > 0$ .

Squaring any positive number gives a positive result.

Therefore:

$\vec{a}.\vec{a} = |\vec{a}|^2 > 0$

For $\vec{a} = 3\hat{i} + 4\hat{j}$ :

Magnitude: $|\vec{a}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$

$\vec{a}.\vec{a} = (3\hat{i} + 4\hat{j}).(3\hat{i} + 4\hat{j})$

$= 9 + 16$

$= 25$

$= 5^2 > 0$

$\vec{a}.\vec{a} = 0$ only when $\vec{a}$ is a zero vector.

$\vec{a}.\vec{a} < 0$ is never possible since squared values are always non-negative.

$\vec{a}.\vec{a} = 1$ only when $|\vec{a}| = 1$ (unit vector), not for all non-zero vectors.

$\vec{a}.\vec{a} > 0$ is always true for any non-zero vector.

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(A) $|\vec a|=14$

(B) $|\vec a|=|\vec b|$

(D) angle between $\vec a$ and $\vec b$ is $\cos^{-1}\left(\dfrac{5}{7}\right)$

Choose the correct answer from the options given below:

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If a⃗\vec{a}a is a non-zero vector, then always

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If $\vec{a}$ is a non-zero vector, then always