Let the volume of a parallelepiped whose coterminous edges are given by $\overrightarrow{\mathrm{u}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}, \overrightarrow{\mathrm{v}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{w}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ be 1 cu. Unit. If $\theta$ be the angle between the edges $\overrightarrow{\mathrm{u}}$ and $\overrightarrow{\mathrm{w}}$, then $\cos \theta$ can be:

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Vector Algebra

Let the volume of a parallelepiped whose coterminous edges are given by

u = \hat{i} + \hat{j} + λ \hat{k}, v = \hat{i} + \hat{j} + 3 \hat{k}

and

w = 2 \hat{i} + \hat{j} + \hat{k}

be 1 cu. Unit. If

θ

be the angle between the edges

u

and

w

, then

cos θ

can be:

Answer: (A)

The volume of the parallelepiped is the magnitude of the scalar triple product of the three edge vectors. Expanding the determinant with rows

(1, 1, λ)

(1, 1, 3)

and

(2, 1, 1)

gives

3 - λ

, so requiring the volume to be one cubic unit forces

∣3 - λ ∣ = 1

, that is

λ = 2

λ = 4

. The angle

θ

between

u = \hat{i} + \hat{j} + λ \hat{k}

and

w = 2 \hat{i} + \hat{j} + \hat{k}

obeys

cos θ = \frac{u \cdot w}{∣ u ∣ ∣ w ∣} = \frac{3 + λ}{2 + λ ^{2} 6}

. For

λ = 2

this is

\frac{5}{6 6} = \frac{5}{6}

, which is not among the choices. For

λ = 4

it becomes

\frac{7}{18 6} = \frac{7}{108} = \frac{7}{6 3}

, the value that

cos θ

can take.

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option