let a be a non-zero vector of magnitude 'a' and is a non-zero scalar, then is a unit vector if Option 1: = -1 Option 2: a = 1/( ) Option 3: = 1 Option 4: a =

Answer: a = 1/( ). Solution: Given a is a non-zero vector with magnitude a (meaning, a, = a) and is a non-zero scalar. A unit vector has magnitude equal to 1. For to be a unit vector:, = 1 When multiplying a vector by a scalar, the magnitude becomes:, =, x, a, =, x a Note:, is used because magnitude is always positive, even if is negative. Setting the magnitude equal to 1:, x a = 1 a = 1/(, ) This condition works for any value of (positive or negative). As long as the magnitude a equals 1/(, ), the vector will be a unit vector. Other options are insufficient: - = -1 or = 1 only work when a = 1 specifically, not as general conditions - a =, gives, =, ^2, which equals 1 only if, = 1 Therefore, the condition is: a = 1/(, )

Algebra

Vector Algebra

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let $\vec{a}$ be a non-zero vector of magnitude ' $a$ ' and $\lambda$ is a non-zero scalar, then $\lambda\vec{a}$ is a unit vector if

$\lambda = -1$

$a = \frac{1}{|\lambda|}$

$\lambda = 1$

$a = |\lambda|$

✅ Correct Option: 2

Given $\vec{a}$ is a non-zero vector with magnitude $a$ (meaning $|\vec{a}| = a$ ) and $\lambda$ is a non-zero scalar.

A unit vector has magnitude equal to 1.

For $\lambda\vec{a}$ to be a unit vector:

$|\lambda\vec{a}| = 1$

When multiplying a vector by a scalar, the magnitude becomes:

$|\lambda\vec{a}| = |\lambda| \times |\vec{a}|$

$|\lambda\vec{a}| = |\lambda| \times a$

Note: $|\lambda|$ is used because magnitude is always positive, even if $\lambda$ is negative.

Setting the magnitude equal to 1:

$|\lambda| \times a = 1$

$a = \frac{1}{|\lambda|}$

This condition works for any value of $\lambda$ (positive or negative). As long as the magnitude $a$ equals $\frac{1}{|\lambda|}$ , the vector $\lambda\vec{a}$ will be a unit vector.

Other options are insufficient:

$\lambda = -1$ or $\lambda = 1$ only work when $a = 1$ specifically, not as general conditions
$a = |\lambda|$ gives $|\lambda\vec{a}| = |\lambda|^2$ , which equals 1 only if $|\lambda| = 1$

Therefore, the condition is: $a = \dfrac{1}{|\lambda|}$

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let a⃗\vec{a}a be a non-zero vector of magnitude 'aaa' and λ\lambdaλ is a non-zero scalar, then λa⃗\lambda\vec{a}λa is a unit vector if

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let $\vec{a}$ be a non-zero vector of magnitude ' $a$ ' and $\lambda$ is a non-zero scalar, then $\lambda\vec{a}$ is a unit vector if