Question

Consider lines $L_1: \overrightarrow{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$ and $L_2: 2\hat{i} + 4\hat{j} + 6\hat{k} + \mu(3\hat{i} + 7\hat{j} + 6\hat{k})$. If point P, Q lies on the lines $L_1$ and $L_2$ respectively, then $\overrightarrow{PQ}$ CANNOT be parallel to

• A. $\hat{i} + \hat{j} + 8\hat{k}$
• B. $\hat{i} + \hat{j} + 2\hat{k}$
• C. $\hat{i} + \hat{j} + 4\hat{k}$
• D. $\hat{i} + \hat{j} + 6\hat{k}$

Answer 1

Answer: (B)

$\hat{i} + \hat{j} + 2\hat{k}$

Answer 2

Let P be a point on $L_1$ and Q be a point on $L_2$. The position vectors are: $\overrightarrow{OP} = (1+2\lambda)\hat{i} + (2+3\lambda)\hat{j} + (3+4\lambda)\hat{k}$ $\overrightarrow{OQ} = (2+3\mu)\hat{i} + (4+7\mu)\hat{j} + (6+6\mu)\hat{k}$

The vector $\overrightarrow{PQ}$ is given by $\overrightarrow{OQ} - \overrightarrow{OP}$: $\overrightarrow{PQ} = [(2+3\mu) - (1+2\lambda)]\hat{i} + [(4+7\mu) - (2+3\lambda)]\hat{j} + [(6+6\mu) - (3+4\lambda)]\hat{k}$ $\overrightarrow{PQ} = (1+3\mu-2\lambda)\hat{i} + (2+7\mu-3\lambda)\hat{j} + (3+6\mu-4\lambda)\hat{k}$

We can rewrite $\overrightarrow{PQ}$ as: $\overrightarrow{PQ} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \mu(3\hat{i} + 7\hat{j} + 6\hat{k}) - \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$

Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ (the difference between the position vectors of points on $L_2$ and $L_1$). Let $\vec{d}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$ (direction vector of $L_1$). Let $\vec{d}_2 = 3\hat{i} + 7\hat{j} + 6\hat{k}$ (direction vector of $L_2$).

So, $\overrightarrow{PQ} = \vec{a} + \mu\vec{d}_2 - \lambda\vec{d}_1$.

The vector $\overrightarrow{PQ}$ can be parallel to a vector $\vec{v}$ if $\overrightarrow{PQ} = k\vec{v}$ for some scalar $k$. This implies $\vec{a} + \mu\vec{d}_2 - \lambda\vec{d}_1 = k\vec{v}$, or $\vec{a} = k\vec{v} - \mu\vec{d}_2 + \lambda\vec{d}_1$. This means that $\vec{a}$ must lie in the span of $\{\vec{v}, \vec{d}_1, \vec{d}_2\}$.

First, let's check if $\vec{a}$ lies in the span of $\{\vec{d}_1, \vec{d}_2\}$. This means checking if $\vec{a} = c_1\vec{d}_1 + c_2\vec{d}_2$ for scalars $c_1, c_2$. $\hat{i} + 2\hat{j} + 3\hat{k} = c_1(2\hat{i} + 3\hat{j} + 4\hat{k}) + c_2(3\hat{i} + 7\hat{j} + 6\hat{k})$ Equating coefficients:

1. $1 = 2c_1 + 3c_2$
2. $2 = 3c_1 + 7c_2$
3. $3 = 4c_1 + 6c_2$

Solving equations 1 and 2: Multiply eq 1 by 3 and eq 2 by 2: $3 = 6c_1 + 9c_2$ $4 = 6c_1 + 14c_2$ Subtracting the first from the second: $1 = 5c_2 \implies c_2 = 1/5$. Substitute $c_2 = 1/5$ into eq 1: $1 = 2c_1 + 3(1/5) \implies 1 = 2c_1 + 3/5 \implies 2/5 = 2c_1 \implies c_1 = 1/5$.

Now check if these values satisfy eq 3: $4c_1 + 6c_2 = 4(1/5) + 6(1/5) = 4/5 + 6/5 = 10/5 = 2$. Since $2 \neq 3$, $\vec{a}$ does NOT lie in the span of $\{\vec{d}_1, \vec{d}_2\}$.

Since $\vec{a}$ is not in the span of $\{\vec{d}_1, \vec{d}_2\}$, for $\vec{a}$ to be in the span of $\{\vec{v}, \vec{d}_1, \vec{d}_2\}$, the vector $\vec{v}$ must NOT be in the span of $\{\vec{d}_1, \vec{d}_2\}$. If $\vec{v}$ were in the span of $\{\vec{d}_1, \vec{d}_2\}$, then the span of $\{\vec{v}, \vec{d}_1, \vec{d}_2\}$ would be the same as the span of $\{\vec{d}_1, \vec{d}_2\}$, which does not contain $\vec{a}$.

Therefore, $\overrightarrow{PQ}$ can be parallel to $\vec{v}$ if and only if $\vec{v}$ is NOT in the plane spanned by $\{\vec{d}_1, \vec{d}_2\}$. The question asks for which vector $\overrightarrow{PQ}$ CANNOT be parallel to. This means we are looking for a vector $\vec{v}$ that IS in the span of $\{\vec{d}_1, \vec{d}_2\}$.

We check which of the given options is a linear combination of $\vec{d}_1$ and $\vec{d}_2$. A vector $\vec{v}$ is in the span of $\vec{d}_1$ and $\vec{d}_2$ if and only if the scalar triple product $[\vec{v}, \vec{d}_1, \vec{d}_2] = 0$.

Let's test option B: $\vec{v}_B = \hat{i} + \hat{j} + 2\hat{k}$. $[\vec{v}_B, \vec{d}_1, \vec{d}_2] = \begin{vmatrix} 1 & 1 & 2 \\ 2 & 3 & 4 \\ 3 & 7 & 6 \end{vmatrix}$ $= 1 \begin{vmatrix} 3 & 4 \\ 7 & 6 \end{vmatrix} - 1 \begin{vmatrix} 2 & 4 \\ 3 & 6 \end{vmatrix} + 2 \begin{vmatrix} 2 & 3 \\ 3 & 7 \end{vmatrix}$ $= 1(18 - 28) - 1(12 - 12) + 2(14 - 9)$ $= 1(-10) - 1(0) + 2(5)$ $= -10 - 0 + 10 = 0$.

Since the scalar triple product is 0, $\vec{v}_B$ is in the span of $\{\vec{d}_1, \vec{d}_2\}$. Thus, $\overrightarrow{PQ}$ CANNOT be parallel to $\hat{i} + \hat{j} + 2\hat{k}$.

For completeness, let's check the other options: A. $\vec{v}_A = \hat{i} + \hat{j} + 8\hat{k}$ $[\vec{v}_A, \vec{d}_1, \vec{d}_2] = \begin{vmatrix} 1 & 1 & 8 \\ 2 & 3 & 4 \\ 3 & 7 & 6 \end{vmatrix} = 1(18-28) - 1(12-12) + 8(14-9) = -10 - 0 + 8(5) = 30 \neq 0$.

C. $\vec{v}_C = \hat{i} + \hat{j} + 4\hat{k}$ $[\vec{v}_C, \vec{d}_1, \vec{d}_2] = \begin{vmatrix} 1 & 1 & 4 \\ 2 & 3 & 4 \\ 3 & 7 & 6 \end{vmatrix} = 1(18-28) - 1(12-12) + 4(14-9) = -10 - 0 + 4(5) = 10 \neq 0$.

D. $\vec{v}_D = \hat{i} + \hat{j} + 6\hat{k}$ $[\vec{v}_D, \vec{d}_1, \vec{d}_2] = \begin{vmatrix} 1 & 1 & 6 \\ 2 & 3 & 4 \\ 3 & 7 & 6 \end{vmatrix} = 1(18-28) - 1(12-12) + 6(14-9) = -10 - 0 + 6(5) = 20 \neq 0$.

Therefore, the only vector that lies in the span of $\vec{d}_1$ and $\vec{d}_2$ is $\hat{i} + \hat{j} + 2\hat{k}$.