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

Concept and Intuition

The condition $\bar a\times\bar b=2(\bar a\times\bar c)$ can be rewritten as $\bar a\times(\bar b-2\bar c)=\bar 0$, which says $\bar b-2\bar c$ is parallel to $\bar a$ — exactly matching the given

$\bar b-2\bar c=\lambda\bar a$. To pin down $\lambda$, take the magnitude of both sides, which only

needs the given lengths $|\bar a|,|\bar b|,|\bar c|$ and the angle between $\bar b$ and $\bar c$.

Step-by-Step Solution

1. From $\bar a\times\bar b=2(\bar a\times\bar c)$: $\bar a\times\bar b-2\bar a\times\bar c=\bar0 \Rightarrow \bar a\times(\bar b-2\bar c)=\bar 0$, confirming $\bar b-2\bar c$ is parallel to $\bar a$ (given as $=\lambda\bar a$).
2. Compute $\bar b\cdot\bar c = |\bar b||\bar c|\cos\theta = 4\cdot1\cdot\dfrac14=1$ (since $\theta=\cos^{-1}(1/4)$).
3. Take magnitudes of $\bar b-2\bar c=\lambda\bar a$: $|\bar b-2\bar c|^2 = |\bar b|^2 - 4(\bar b\cdot\bar c) + 4|\bar c|^2 = 16 - 4(1) + 4(1) = 16.$ …

Concept and Intuition

$\vec u\times\vec a=\vec v\times\vec a\iff(\vec u-\vec v)\times\vec a=\vec 0\iff \vec u-\vec v$ is parallel to $\vec a$ (or zero). Applying this twice — once for the $a$-condition, once for the $b$-condition — gives two linear constraints on $r$ that combine to a unique vector when $a,b$ are non-parallel.

Step-by-Step Solution

1. Given $\bar a=\bar i+\bar j=(1,1,0)$, $\bar b=2\bar j-\bar k=(0,2,-1)$ — not parallel (not scalar multiples).
2. From $\bar r\times\bar a=\bar b\times\bar a$: $(\bar r-\bar b)\times\bar a=\bar 0\Rightarrow \bar r-\bar b=t\,\bar a$ for some scalar $t$, i.e. $\bar r=\bar b+t\bar a$.
3. From $\bar r\times\bar b=\bar a\times\bar b$: $(\bar r-\bar a)\times\bar b=\bar 0\Rightarrow \bar r=\bar a+s\bar b$ for some scalar $s$.
4. Equating: $\bar b+t\bar a=\bar a+s\bar b\Rightarrow(t-1)\bar a=(s-1)\bar b$. Since $\bar a,\bar b$ are linearly independent, this forces $t-1=0$ and $s-1=0$, i.e. $t=s=1$. …

Concept and Intuition

A plane parallel to two given vectors has those vectors' cross product as its normal. Once the normal is known, the plane equation follows immediately from the point-normal form.

Step-by-Step Solution

1. Direction vectors: $\vec p=(2,1,1)$, $\vec q=(1,-1,1)$.
2. Normal $\vec n = \vec p\times\vec q = \begin{vmatrix}\hat i&\hat j&\hat k\\2&1&1\\1&-1&1\end{vmatrix} = \hat i(1\cdot1-1\cdot(-1)) - \hat j(2\cdot1-1\cdot1) + \hat k(2\cdot(-1)-1\cdot1)$.
3. $= \hat i(1+1) - \hat j(2-1) + \hat k(-2-1) = 2\hat i-\hat j-3\hat k$.
4. Plane through $(3,2,6)$ with normal $(2,-1,-3)$: $2(x-3)-1(y-2)-3(z-6)=0$. …

Concept and Intuition

Given the two defining cross-product relations, first extract any perpendicularity facts from them (a cross product is always perpendicular to both its factors), then substitute $\vec c=\vec a\times\vec b$ into the target expression and simplify using the standard vector triple product identity.

Step-by-Step Solution

1. Since $\vec c=\vec a\times\vec b$, $\vec c$ is perpendicular to $\vec a$ (and to $\vec b$).
2. Since $\vec a=\vec b\times\vec c$, $\vec a$ is perpendicular to $\vec b$ (and to $\vec c$).
3. Combined: $\vec a\perp\vec b$, i.e. $\vec a\cdot\vec b=0$.
4. Compute $\vec a\times\vec c=\vec a\times(\vec a\times\vec b)$ using the triple product identity $\vec X\times(\vec Y\times\vec Z)=\vec Y(\vec X\cdot\vec Z)-\vec Z(\vec X\cdot\vec Y)$ with $X=Y=\vec a$, $Z=\vec b$: …

Concept and Intuition

A plane spanned by two vectors has normal equal to their cross product. The line common to two planes is perpendicular to both normals, i.e., parallel to the cross product of the two normals.

Step-by-Step Solution

1. Normal to $\pi_1$ (spanned by $\bar i+\bar j$, $\bar j+\bar k$): $(\bar i+\bar j)\times(\bar j+\bar k)=(1,-1,1)$.
2. Normal to $\pi_2$ (spanned by $\bar i-\bar j$, $\bar i+\bar j-\bar k$): $(\bar i-\bar j)\times(\bar i+\bar j-\bar k)=(1,1,2)$.
3. Direction of the line of intersection: $(1,-1,1)\times(1,1,2)=(-3,-1,2)$. …

Concept and Intuition

"Parallel to $\bar c$" pins $\bar c$ down to a single free scalar $\lambda$ along the fixed direction $\bar a+\bar b$. The dot-product condition then becomes an ordinary quadratic in $\lambda$, typically with two roots — the problem asks specifically for whichever root gives the smaller magnitude.

Step-by-Step Solution

1. $\bar a=\bar i-\bar j+2\bar k,\ \bar b=-\bar i+2\bar j$, so $\bar a+\bar b=\bar j+2\bar k=(0,1,2)$.
2. Since $\bar c\parallel(\bar a+\bar b)$, write $\bar c=\lambda(0,1,2)=(0,\lambda,2\lambda)$.
3. $\bar c+\bar a=(1,\lambda-1,2\lambda+2)$; $\bar c+\bar b=(-1,\lambda+2,2\lambda)$.
4. $(\bar c+\bar a)\cdot(\bar c+\bar b)=(1)(-1)+(\lambda-1)(\lambda+2)+(2\lambda+2)(2\lambda)$ $=-1+(\lambda^2+\lambda-2)+(4\lambda^2+4\lambda)=5\lambda^2+5\lambda-3$.
5. Set equal to 7: $5\lambda^2+5\lambda-3=7 \Rightarrow 5\lambda^2+5\lambda-10=0 \Rightarrow \lambda^2+\lambda-2=0 \Rightarrow (\lambda-1)(\lambda+2)=0$. …

Concept and Intuition

Any vector perpendicular to two given (non-parallel) vectors must lie along their cross product — that's the whole geometric meaning of the cross product. So $\vec x=k(\vec a\times\vec b)$ for some scalar $k$, and the remaining conditions (magnitude, angle with $\vec j$) just pin down $k$.

Step-by-Step Solution

1. Compute $\vec a\times\vec b$ with $\vec a=3\vec i+2\vec j+2\vec k,\ \vec b=18\vec i-22\vec j-5\vec k$:

2. Factor out the common factor $17$: $\vec a\times\vec b=17(2,3,-6)$. Note $|(2,3,-6)|=\sqrt{4+9+36}=7$.
3. So $\vec x=k(2,3,-6)$ for some real $k$.
4. The angle with $\vec j=(0,1,0)$ is obtuse $\iff \vec x\cdot\vec j<0 \iff$ the $y$-component of $\vec x$ is negative $\iff k<0$. …

Concept and Intuition

A vector perpendicular to the plane containing $\bar a$ and $\bar b$ must be a scalar multiple of $\bar a\times\bar b$ (the plane's normal direction) — that's the defining property of the cross product. The extra condition $\bar d\cdot\bar c=2$ then pins down exactly which multiple $\bar d$ is.

Step-by-Step Solution

1. $\bar a=(2,-1,2),\ \bar b=(1,-2,2)$. Compute $\bar a\times\bar b = \big((-1)(2)-(2)(-2),\ (2)(1)-(2)(2),\ (2)(-2)-(-1)(1)\big) = (-2+4,\ 2-4,\ -4+1)=(2,-2,-3)$.
2. Since $\bar d$ is perpendicular to the plane of $\bar a,\bar b$, $\bar d=\lambda(2,-2,-3)$ for some scalar $\lambda$.
3. $\bar c=(2,-2,-1)$. $\bar d\cdot\bar c=\lambda\left[2(2)+(-2)(-2)+(-3)(-1)\right]=\lambda(4+4+3)=11\lambda=2\Rightarrow\lambda=\dfrac{2}{11}$.
4. So $\bar d=\left(\dfrac{4}{11},-\dfrac{4}{11},-\dfrac{6}{11}\right)$.
5. $\bar d-\bar c=\left(\dfrac{4}{11}-2,\ -\dfrac{4}{11}+2,\ -\dfrac{6}{11}+1\right)=\left(-\dfrac{18}{11},\dfrac{18}{11},\dfrac{5}{11}\right)$. …