You visited us 1 times! Enjoying our articles? Unlock Full Access!

Secant of a Curve y =f(x)

Find the shor...

Question

Find the shortest distance and the vector equation of the line of shortest distance between the lines given by:
$\to r = (3 i + 8 j + 3 k) + λ (3 i - j + k)$
$\to r = (- 3 i - 7 j + 6 k) + μ (- 3 i + 2 j + 4 k)$

3 \sqrt{30}

units

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\sqrt{90}

units

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 \sqrt{30}

units

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 \sqrt{10}

units

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $3 \sqrt{30}$ units
We have $_{1} = 3^i + 8^j + 3^k$ $_{1} = 3^i -^j +^k$
$_{2} = - 3^i - 7^j + 6^k$ and $_{2} = - 3^i + 2^j + 4^k$
Now, $b_{1} \times b_{2} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 3 & - 1 & 1 - 3 & 2 & 4 \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 6^i - 15^j + 3^k$
$∣ b_{1} \times b_{2} ∣= \sqrt{6^{2} + 15^{2} + 3^{2}} = 3 \sqrt{30}$
$a_{1} - a_{2} = 6^i + 15^j - 3^k$
Therefore shortest distance between the given lines is

$d =∣ \frac{(a_{1} - a_{2}) \times (b_{1} \times b_{2})}{∣ b_{1} \times b_{2} ∣} ∣= \frac{6^{2} + 15^{2} + 3^{2}}{3 \sqrt{30}} = 3 \sqrt{30}$

Suggest Corrections

Similar questions

\to r = (3 i + 8 j + k) + λ (3 i - j + k)

\to r = (- 3 i - 7 j + 6 k) + μ (3 i 2 j + 4 k)

\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} + λ (3 \hat{i} - \hat{j} + \hat{k}) and \vec{r} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} + μ (- 3 \hat{i} + 2 \hat{j} + 4 \hat{k})

\vec{r} = (3 \hat{i} + 5 \hat{j} + 7 \hat{k}) + λ (\hat{i} - 2 \hat{j} + 7 \hat{k}) and \vec{r} = - \hat{i} - \hat{j} - \hat{k} + μ (7 \hat{i} - 6 \hat{j} + \hat{k})

\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) and \vec{r} = (2 \hat{i} + 4 \hat{j} + 5 \hat{k}) + μ (3 \hat{i} + 4 \hat{j} + 5 \hat{k})

\vec{r} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - t) \hat{k} and \vec{r} = (s + 1) \hat{i} + (2 s - 1) \hat{j} - (2 s + 1) \hat{k}

\vec{r} = (λ - 1) \hat{i} + (λ + 1) \hat{j} - (1 + λ) \hat{k} and \vec{r} = (1 - μ) \hat{i} + (2 μ - 1) \hat{j} + (μ + 2) \hat{k}

\vec{r} = (2 \hat{i} - \hat{j} - \hat{k}) + λ (2 \hat{i} - 5 \hat{j} + 2 \hat{k}) and, \vec{r} = (\hat{i} + 2 \hat{j} + \hat{k}) + μ (\hat{i} - \hat{j} + \hat{k})

\vec{r} = (\hat{i} + \hat{j}) + λ (2 \hat{i} - \hat{j} + \hat{k}) and, \vec{r} = 2 \hat{i} + \hat{j} - \hat{k} + μ (3 \hat{i} - 5 \hat{j} + 2 \hat{k})

\vec{r} = (8 + 3 λ) \hat{i} - (9 + 16 λ) \hat{j} + (10 + 7 λ) \hat{k}

\vec{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + μ (3 \hat{i} + 8 \hat{j} - 5 \hat{k})

\to r

\to r \times \to b = \to r \times \to c

\to b = ˆ i + 2 ˆ j + ˆ k

\to c = 3 ˆ i + 2 ˆ k

r = 3 i + 8 j + 7 k + λ (1 - 2 j + k)

r = - i - j - k + μ (7 i - 6 j + k)

\to r = (4 i - j) + λ (i + 2 j - 3 k)

\to r = (i - j + 2 k) + μ (2 i + 4 j - 5 k)

Join BYJU'S Learning Program