Q.Find $\lambda$ and $\mu$ if $(2\hat{i}+6\hat{j}+27\hat{k})\times(\hat{i}+\lambda\hat{j}+\mu\hat{k})=\vec{0}$.

We are given three vectors $\vec{a}, \vec{b}, \vec{c}$ with $\vec{a} \neq \vec{0}$, and the conditions:

Also, $\vec{b} - 2\vec{c} = \lambda \vec{a}$. We need $\lambda$.

Concept and intuition:

The equation $\vec{a} \times \vec{b} = 2 \vec{a} \times \vec{c}$ can be rewritten as $\vec{a} \times (\vec{b} - 2\vec{c}) = \vec{0}$. This tells us that $\vec{b} - 2\vec{c}$ is parallel to $\vec{a}$, which is exactly the given relation $\vec{b} - 2\vec{c} = \lambda \vec{a}$. So the cross-product condition already confirms the direction; the unknown is the scalar $\lambda$. To find $\lambda$, we use the magnitudes: we know $|\vec{a}|=1$, so $|\vec{b} - 2\vec{c}| = |\lambda|$. Compute $|\vec{b} - 2\vec{c}|^2$ using the given magnitudes and the cross-product magnitude $|\vec{b} \times \vec{c}|$ to find the dot product $\vec{b} \cdot \vec{c}$, then solve for $\lambda$.

Step-by-step solution:

1. Rewrite the cross-product condition.

Since $\vec{a} \neq \vec{0}$, this means $\vec{b} - 2\vec{c}$ is parallel to $\vec{a}$. So there exists a scalar $\lambda$ such that $\vec{b} - 2\vec{c} = \lambda \vec{a}$. This is consistent with the problem statement.

2. Express $|\vec{b} - 2\vec{c}|^2$ in terms of known quantities.

We know $|\vec{b}| = 4$ and $|\vec{c}| = 1$, so:

3. Find $\vec{b} \cdot \vec{c}$ using the given $|\vec{b} \times \vec{c}| = \sqrt{15}$. The identity relating cross product magnitude and dot product is:

Substitute known values:

Hence:

4. Determine the sign of $\vec{b} \cdot \vec{c}$. We also have the relation $\vec{b} - 2\vec{c} = \lambda \vec{a}$. Take the dot product of both sides with $\vec{c}$:

The left side is $\vec{b} \cdot \vec{c} - 2|\vec{c}|^2 = \vec{b} \cdot \vec{c} - 2$.

The right side is $\lambda (\vec{a} \cdot \vec{c})$. We don’t know $\vec{a} \cdot \vec{c}$ directly, but we can also take the dot product with $\vec{b}$:

Left side: $|\vec{b}|^2 - 2(\vec{b} \cdot \vec{c}) = 16 - 2(\vec{b} \cdot \vec{c})$.

However, a simpler approach: Since $\vec{b} - 2\vec{c}$ is parallel to $\vec{a}$, its magnitude squared is $|\lambda|^2 |\vec{a}|^2 = \lambda^2$ (since $|\vec{a}|=1$). So:

If $\vec{b} \cdot \vec{c} = +1$, then $\lambda^2 = 20 - 4 = 16 \implies \lambda = \pm 4$.

If $\vec{b} \cdot \vec{c} = -1$, then $\lambda^2 = 20 + 4 = 24 \implies \lambda = \pm \sqrt{24} = \pm 2\sqrt{6}$, which is not among the options (2, 1, -1, -4). So $\vec{b} \cdot \vec{c}$ must be $+1$, and $\lambda = \pm 4$.

5. Choose the correct sign for $\lambda$. We have $\vec{b} - 2\vec{c} = \lambda \vec{a}$. Take the cross product with $\vec{a}$ on both sides:

which is automatically satisfied. That doesn’t fix the sign.

Instead, consider the dot product with $\vec{a}$:

So $\lambda = \vec{a} \cdot \vec{b} - 2(\vec{a} \cdot \vec{c})$. We don’t know these dot products directly, but we can use the cross-product condition again: $\vec{a} \times \vec{b} = 2 \vec{a} \times \vec{c}$ implies $\vec{a} \times (\vec{b} - 2\vec{c}) = 0$, which we already used. To determine sign, note that from $\vec{b} - 2\vec{c} = \lambda \vec{a}$, taking magnitude gives $|\lambda| = 4$. Now, also take the dot product with $\vec{b}$:

With $\vec{b} \cdot \vec{c} = 1$, left side is $14$. So $\lambda (\vec{a} \cdot \vec{b}) = 14$. Similarly, dot with $\vec{c}$:

So $\vec{a} \cdot \vec{c} = -1/\lambda$. Since $|\vec{a} \cdot \vec{c}| \le |\vec{a}||\vec{c}| = 1$, this is fine. Now, we also have the identity:

But we already used that. To decide sign, note that if $\lambda = +4$, then $\vec{a} \cdot \vec{c} = -1/4$ and $\vec{a} \cdot \vec{b} = 14/4 = 3.5$. If $\lambda = -4$, then $\vec{a} \cdot \vec{c} = 1/4$ and $\vec{a} \cdot \vec{b} = -3.5$. Both are possible geometrically. However, we have one more unused condition: the original cross-product relation $\vec{a} \times \vec{b} = 2 \vec{a} \times \vec{c}$ implies that the vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar? Actually, $\vec{a} \times \vec{b}$ and $\vec{a} \times \vec{c}$ are both perpendicular to $\vec{a}$, so the equality says their components perpendicular to $\vec{a}$ are related. But we already used that to get parallelism.

A decisive check: Use the vector triple product identity. Since $\vec{b} - 2\vec{c} = \lambda \vec{a}$, cross with $\vec{b}$:

Left side: $\vec{b} \times \vec{b} - 2(\vec{c} \times \vec{b}) = 0 + 2(\vec{b} \times \vec{c}) = 2(\vec{b} \times \vec{c})$.

So $2(\vec{b} \times \vec{c}) = \lambda (\vec{a} \times \vec{b})$. But we also have $\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})$. Substitute:

Take magnitudes: $|\vec{b} \times \vec{c}| = |\lambda| |\vec{a} \times \vec{c}|$. We know $|\vec{b} \times \vec{c}| = \sqrt{15}$. Also, $|\vec{a} \times \vec{c}| = \sqrt{|\vec{a}|^2 |\vec{c}|^2 - (\vec{a} \cdot \vec{c})^2} = \sqrt{1 - (\vec{a} \cdot \vec{c})^2}$.

If $\lambda = +4$, then $\vec{a} \cdot \vec{c} = -1/4$, so $|\vec{a} \times \vec{c}| = \sqrt{1 - 1/16} = \sqrt{15/16} = \sqrt{15}/4$. Then $|\lambda| |\vec{a} \times \vec{c}| = 4 \cdot (\sqrt{15}/4) = \sqrt{15}$, which matches.

If $\lambda = -4$, then $\vec{a} \cdot \vec{c} = 1/4$, same magnitude, so also matches. So magnitude alone doesn’t decide sign.

However, note the equation $\vec{b} \times \vec{c} = \lambda (\vec{a} \times \vec{c})$ is a vector equation. If $\lambda = +4$, then $\vec{b} \times \vec{c}$ is in the same direction as $\vec{a} \times \vec{c}$. If $\lambda = -4$, it’s opposite. Both are possible given the data? Let’s check consistency with the given cross-product condition: $\vec{a} \times \vec{b} = 2 \vec{a} \times \vec{c}$. Cross both sides with $\vec{a}$? Alternatively, use the fact that $\vec{b} - 2\vec{c} = \lambda \vec{a}$ implies $\vec{b} = \lambda \vec{a} + 2\vec{c}$. Substitute into $\vec{a} \times \vec{b} = \vec{a} \times (\lambda \vec{a} + 2\vec{c}) = \lambda (\vec{a} \times \vec{a}) + 2(\vec{a} \times \vec{c}) = 2(\vec{a} \times \vec{c})$. This holds for any $\lambda$! So the cross-product condition gives no restriction on $\lambda$ beyond the parallelism. …

Since $\vec a\times\vec b = 2(\vec a\times\vec c)$, we have $\vec a\times(\vec b-2\vec c)=\vec 0$, so $\vec b-2\vec c$ is parallel to $\vec a$, consistent with $\vec b-2\vec c=\lambda\vec a$.

Take magnitudes (with $|\vec a|=1$):

Find $\vec b\cdot\vec c$ from the cross-product magnitude: …

From $\vec r\times\vec a=\vec b\times\vec a$: $(\vec r-\vec b)\times\vec a=\vec 0$, so $\vec r-\vec b=\lambda\vec a$, i.e. $\vec r=\vec b+\lambda\vec a$.

From $\vec r\times\vec b=\vec a\times\vec b$: $(\vec r-\vec a)\times\vec b=\vec 0$, so $\vec r=\vec a+\mu\vec b$.

Equating: $\vec b+\lambda\vec a=\vec a+\mu\vec b\implies(\lambda-1)\vec a=(\mu-1)\vec b$. As $\vec a$ and $\vec b$ are not parallel, $\lambda=1,\mu=1$. …

Step 1 — Why this identity works

With $\theta$ the angle between the vectors,

Squaring and adding, and using $\sin^2\theta + \cos^2\theta = 1$:

Step 2 — Substitute the data

Step 3 — Solve …