JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Vector Algebra

If vectors

a_{1} = x \hat{i} - \hat{j} + \hat{k}

and

a_{2} = \hat{i} + y \hat{j} + z \hat{k}

are collinear, then a possible unit vector parallel to the vector

x \hat{i} + y \hat{j} + z \hat{k}

is :

Answer: (C)

Solution

Collinearity means

a_{1} = λ a_{2}

for some scalar

λ

, so comparing components,

x = λ, - 1 = λ y, 1 = λ z

which give

y = - \frac{1}{λ}

and

z = \frac{1}{λ}

. Hence

x \hat{i} + y \hat{j} + z \hat{k} = λ \hat{i} - \frac{1}{λ} \hat{j} + \frac{1}{λ} \hat{k}

and the choice

λ = 1

gives

\hat{i} - \hat{j} + \hat{k}

, whose magnitude is

3

. A possible unit vector parallel to

x \hat{i} + y \hat{j} + z \hat{k}

\frac{1}{3} (\hat{i} - \hat{j} + \hat{k})

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option