If the shortest distance between the line \(\overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}+3 \mathrm{k})+\lambda(\hat{\mathrm{i}}-\mathrm{a} \hat{\mathrm{j}})\) and \(\overrightarrow{\mathrm{r}}=(-\hat{\mathrm{j}}+2 \mathrm{k})+\mu(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\mathrm{k})\) is \(\sqrt{\frac{2}{3}}\), then the integral value of a is equal to ______.

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Three Dimensional Geometry

If the shortest distance between the line

r = (- \hat{i} + 3 k) + λ (\hat{i} - a \hat{j})

and

r = (- \hat{j} + 2 k) + μ (\hat{i} - \hat{j} + k)

\frac{2}{3}

, then the integral value of a is equal to ______.

Answer: 2

For the two lines, identify:

a_{1} = (- 1, 0, 3)

b_{1} = (1, - a, 0)

;

a_{2} = (0, - 1, 2)

b_{2} = (1, - 1, 1)

.

The shortest distance is

d = \frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣} .

a_{2} - a_{1} = (1, - 1, - 1)

b_{1} \times b_{2} = \hat{i} 11 \hat{j} - a - 1 \hat{k} 01 = (- a) \hat{i} - \hat{j} + (a - 1) \hat{k}

∣ b_{1} \times b_{2} ∣ = a^{2} + 1 + (a - 1)^{2} = 2 a^{2} - 2 a + 2

(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = - a + 1 - (a - 1) = 2 - 2 a

\frac{∣2 - 2 a ∣}{2 a ^{2} - 2 a + 2} = \frac{2}{3} .

Squaring:

\frac{4 ( 1 - a ) ^{2}}{2 a ^{2} - 2 a + 2} = \frac{2}{3}

, giving

6 (1 - a)^{2} = 2 a^{2} - 2 a + 2

.

Simplifying:

4 a^{2} - 10 a + 4 = 0

, so

a = 2

a = \frac{1}{2}

. The integral value is

a = 2

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Integer answer