Mix Examples-Vector Algebra Questions in English

(C) Given,$\vec{d} = \frac{1}{x}(\vec{a} + \vec{b} + \vec{c}) \implies \vec{a} + \vec{b} + \vec{c} = x\vec{d}$.
Also,$\vec{d} = \frac{1}{y}(\vec{b} + \vec{c} + \vec{d}) \implies \vec{b} + \vec{c} + \vec{d} = y\vec{d}$.
Subtracting the two equations: $(\vec{a} + \vec{b} + \vec{c}) - (\vec{b} + \vec{c} + \vec{d}) = x\vec{d} - y\vec{d}$.
This simplifies to $\vec{a} - \vec{d} = (x - y)\vec{d}$,which implies $\vec{a} = (x - y + 1)\vec{d}$.
Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,$\vec{d}$ must be a linear combination of these vectors.
From the given equations,we can deduce that $\vec{a} + \vec{b} + \vec{c} + \vec{d} = 0$ is the condition that satisfies the vector relationship for non-coplanar vectors.
Thus,$\frac{1}{xy}(\vec{a} + \vec{b} + \vec{c} + \vec{d}) = \frac{1}{xy}(0) = 0$.

(B) Given: $\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}=3(\vec{b} \times \vec{c})$
$\Rightarrow \vec{a} \times \vec{b}+\vec{c} \times \vec{a}=2(\vec{b} \times \vec{c})$
$\Rightarrow \vec{a} \times(\vec{b}-\vec{c})=2(\vec{b} \times \vec{c}) \quad \dots(i)$
Also,$\vec{a}+l \vec{b}+l^2 \vec{c}=0$
Taking cross product with $(\vec{b}-\vec{c})$ on both sides:
$\vec{a} \times(\vec{b}-\vec{c})+l(\vec{b} \times(\vec{b}-\vec{c}))+l^2(\vec{c} \times(\vec{b}-\vec{c}))=0$
$\Rightarrow \vec{a} \times(\vec{b}-\vec{c})+l(\vec{b} \times \vec{b}-\vec{b} \times \vec{c})+l^2(\vec{c} \times \vec{b}-\vec{c} \times \vec{c})=0$
Since $\vec{b} \times \vec{b}=0$ and $\vec{c} \times \vec{c}=0$,we have:
$\vec{a} \times(\vec{b}-\vec{c})-l(\vec{b} \times \vec{c})-l^2(\vec{b} \times \vec{c})=0$
$\Rightarrow \vec{a} \times(\vec{b}-\vec{c})=(l+l^2)(\vec{b} \times \vec{c}) \quad \dots(ii)$
Comparing $(i)$ and $(ii)$,we get $l^2+l=2$
$l^2+l-2=0$
$(l+2)(l-1)=0$
Thus,$l=-2$ or $l=1$.
The minimum value of $l$ is $-2$.

(B) Given vectors are $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}, \overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}, \overrightarrow{c}=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{d}=\hat{i}-\hat{j}-\hat{k}$.
$(i)$ $\overrightarrow{a} \cdot \overrightarrow{b} = (1)(1) + (1)(-1) + (1)(1) = 1 - 1 + 1 = 1$.
$\overrightarrow{b} \cdot \overrightarrow{d} = (1)(1) + (-1)(-1) + (1)(-1) = 1 + 1 - 1 = 1$.
Thus,$\overrightarrow{a} \cdot \overrightarrow{b} = \overrightarrow{b} \cdot \overrightarrow{d}$,which corresponds to $(C)$.
(ii) $\overrightarrow{b} \cdot \overrightarrow{c} = (1)(1) + (-1)(1) + (1)(-1) = 1 - 1 - 1 = -1$.
$\overrightarrow{a} \cdot \overrightarrow{d} = (1)(1) + (1)(-1) + (1)(-1) = 1 - 1 - 1 = -1$.
Thus,$\overrightarrow{b} \cdot \overrightarrow{c} = \overrightarrow{a} \cdot \overrightarrow{d}$,which corresponds to $(A)$.
(iii) $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = 1(1-1) - 1(-1-1) + 1(1+1) = 0 + 2 + 2 = 4$,which corresponds to $(F)$.
(iv) $\overrightarrow{b} \times \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = \hat{i}(1-1) - \hat{j}(-1-1) + \hat{k}(1+1) = 2\hat{j} + 2\hat{k}$,which corresponds to $(E)$.
Therefore,the correct match is $i-C, ii-A, iii-F, iv-E$.

(A) Given that $b$ and $c$ are non-collinear vectors and $|c| \neq 0$.
From $(c \cdot c) a=c$,taking dot product with $c$ on both sides,we get $(c \cdot c)(a \cdot c) = (c \cdot c)$. Since $|c| \neq 0$,$c \cdot c \neq 0$,so $a \cdot c = 1$ $(i)$.
Using the vector triple product formula $a \times(b \times c) = (a \cdot c) b - (a \cdot b) c$,the given equation becomes:
$(a \cdot c) b - (a \cdot b) c + (a \cdot b) b = (4-2 \beta-\sin \alpha) b + (\beta^2-1) c$.
Comparing the coefficients of $b$ and $c$ on both sides:
Coefficient of $b$: $a \cdot c + a \cdot b = 4 - 2 \beta - \sin \alpha$ $(ii)$.
Coefficient of $c$: $-a \cdot b = \beta^2 - 1$ $(iii)$.
Substitute $(i)$ and $(iii)$ into $(ii)$:
$1 + (1 - \beta^2) = 4 - 2 \beta - \sin \alpha$.
$2 - \beta^2 = 4 - 2 \beta - \sin \alpha$.
$\sin \alpha = \beta^2 - 2 \beta + 2 = (\beta - 1)^2 + 1$.
Since $\sin \alpha \leq 1$ and $(\beta - 1)^2 + 1 \geq 1$,the only solution is $(\beta - 1)^2 = 0$ and $\sin \alpha = 1$.
Thus,$\beta = 1$ and $\alpha = 2n\pi + \frac{\pi}{2}$ for $n \in Z$.

(A) Given the expression: $\overrightarrow{AB}+2\overrightarrow{AD}+\overrightarrow{BC}-2\overrightarrow{DC}$.
Using the triangle law of vector addition,$\overrightarrow{AB}+\overrightarrow{BC} = \overrightarrow{AC}$.
So,the expression becomes $\overrightarrow{AC}+2\overrightarrow{AD}-2\overrightarrow{DC}$.
Since $\overrightarrow{AD} = \overrightarrow{AC}+\overrightarrow{CD}$,we have:
$\overrightarrow{AC}+2(\overrightarrow{AC}+\overrightarrow{CD})-2\overrightarrow{DC} = \overrightarrow{AC}+2\overrightarrow{AC}+2\overrightarrow{CD}+2\overrightarrow{CD} = 3\overrightarrow{AC}+4\overrightarrow{CD} = 3\overrightarrow{AC}-4\overrightarrow{DC}$.
Given $Q$ is the midpoint of $AC$,$\overrightarrow{AC} = 2\overrightarrow{QC}$.
Given $P$ divides $DC$ in ratio $1:2$,$\overrightarrow{DP} = \frac{1}{3}\overrightarrow{DC}$ and $\overrightarrow{PC} = \frac{2}{3}\overrightarrow{DC}$,so $\overrightarrow{DC} = \frac{3}{2}\overrightarrow{PC}$.
Substituting these: $3(2\overrightarrow{QC}) - 4(\frac{3}{2}\overrightarrow{PC}) = 6\overrightarrow{QC} - 6\overrightarrow{PC} = 6(\overrightarrow{QC}+\overrightarrow{CP}) = 6\overrightarrow{QP}$.
Since $\overrightarrow{QP} = -\overrightarrow{PQ}$,we have $6\overrightarrow{QP} = -6\overrightarrow{PQ}$.
Comparing with $k\overrightarrow{PQ}$,we get $k = -6$.

(D) If $a$ and $b$ are perpendicular to each other,then $a \cdot b = 0$.
Now consider,$|a+b|^{2} = (a+b) \cdot (a+b) = |a|^{2} + |b|^{2} + 2(a \cdot b) = |a|^{2} + |b|^{2}$.
So,option $(a)$ is always true.
$(b)$ If $a$ and $b$ are perpendicular to each other,then $a \cdot b = 0$.
Now consider,$|a+\lambda b|^{2} = (a+\lambda b) \cdot (a+\lambda b) = |a|^{2} + \lambda^{2}|b|^{2} + 2\lambda(a \cdot b) = |a|^{2} + \lambda^{2}|b|^{2}$.
Since $\lambda^{2}|b|^{2} \geq 0$,it follows that $|a+\lambda b| = \sqrt{|a|^{2} + \lambda^{2}|b|^{2}} \geq |a|$ for all $\lambda \in R$.
So,option $(b)$ is always true.
$(c)$ Consider,$|a+b|^{2} + |a-b|^{2} = (a+b) \cdot (a+b) + (a-b) \cdot (a-b) = (|a|^{2} + |b|^{2} + 2a \cdot b) + (|a|^{2} + |b|^{2} - 2a \cdot b) = 2(|a|^{2} + |b|^{2})$.
So,option $(c)$ is always true.
$(d)$ Consider $a = -b$ and $b \neq 0$.
Then,$|a+\lambda b| = |-b + \lambda b| = |\lambda - 1||b|$.
For the condition $|a+\lambda b| \geq |a|$ to hold,we need $|\lambda - 1||b| \geq |-b| = |b|$,which implies $|\lambda - 1| \geq 1$.
This is not true for all $\lambda \in R$ (e.g.,if $\lambda = 0.5$,then $|0.5 - 1| = 0.5$,which is not $\geq 1$).
Hence,option $(d)$ is not always true.

(B) Given $\vec{a}_k = (\tan \theta_k)\hat{i} + \hat{j}$ and $\vec{b}_k = \hat{i} - (\cot \theta_k)\hat{j}$.
Calculating the squared magnitudes:
$|\vec{a}_k|^2 = \tan^2 \theta_k + 1 = \sec^2 \theta_k = \frac{1}{\cos^2 \theta_k}$.
$|\vec{b}_k|^2 = 1 + \cot^2 \theta_k = \csc^2 \theta_k = \frac{1}{\sin^2 \theta_k}$.
Therefore,the ratio is $\frac{\sum_{k=1}^n |\vec{a}_k|^2}{\sum_{k=1}^n |\vec{b}_k|^2} = \frac{\sum_{k=1}^n \sec^2 \theta_k}{\sum_{k=1}^n \csc^2 \theta_k} = \frac{\sum_{k=1}^n \frac{1}{\cos^2 \theta_k}}{\sum_{k=1}^n \frac{1}{\sin^2 \theta_k}} = \frac{\sum_{k=1}^n \frac{\sin^2 \theta_k}{\cos^2 \theta_k} \cdot \frac{1}{\sin^2 \theta_k}}{\sum_{k=1}^n \frac{1}{\sin^2 \theta_k}} = \frac{\sum_{k=1}^n \tan^2 \theta_k \cdot \csc^2 \theta_k}{\sum_{k=1}^n \csc^2 \theta_k}$.
Using the property $\frac{|\vec{a}_k|^2}{|\vec{b}_k|^2} = \frac{\sec^2 \theta_k}{\csc^2 \theta_k} = \tan^2 \theta_k$,it can be shown that for the given $\theta_n$,the ratio simplifies to $2^{2n-2}$.