Let O be the origin and the position vector of A and B be 2hati+2hatj+hatk$2\hat{i}+2\hat{j}+\hat{k}$ and 2hati+4hatj+4hatk$2\hat{i}+4\hat{j}+4\hat{k}$ respectively. If the internal bisector of angle AOB$\angle AOB$ meets the line AB at C, then the length of OC is
A.frac23sqrt31$\frac{2}{3}\sqrt{31}$
B.frac23sqrt34$\frac{2}{3}\sqrt{34}$
C.frac34sqrt34$\frac{3}{4}\sqrt{34}$
D.frac32sqrt31$\frac{3}{2}\sqrt{31}$
Solution
### Related Formula
textInternal Angle Bisector Theorem: fracACCB = frac|vecOA||vecOB|$$\text{Internal Angle Bisector Theorem: } \frac{AC}{CB} = \frac{|\vec{OA}|}{|\vec{OB}|}$$textSection Formula: vecOC = fracmvecOB + nvecOAm+n$$\text{Section Formula: } \vec{OC} = \frac{m\vec{OB} + n\vec{OA}}{m+n}$$
### Core Logic
Find the magnitudes of the position vectors vecOA$\vec{OA}$ and vecOB$\vec{OB}$:
|vecOA| = sqrt2^2 + 2^2 + 1^2 = sqrt4+4+1 = sqrt9 = 3$$|\vec{OA}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3$$|vecOB| = sqrt2^2 + 4^2 + 4^2 = sqrt4+16+16 = sqrt36 = 6$$|\vec{OB}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4+16+16} = \sqrt{36} = 6$$
According to the internal angle bisector theorem in Delta AOB$\Delta AOB$, the point C divides the segment AB in the ratio of the adjacent sides:
fracACCB = frac|vecOA||vecOB| = frac36 = frac12$$\frac{AC}{CB} = \frac{|\vec{OA}|}{|\vec{OB}|} = \frac{3}{6} = \frac{1}{2}$$Vector Angle Bisector
### Step 1: Apply Section Formula
Using the section formula to find the position vector of C, dividing AB internally in ratio m:n = 1:2$m:n = 1:2$:
vecOC = frac1(vecOB) + 2(vecOA)1 + 2$$\vec{OC} = \frac{1(\vec{OB}) + 2(\vec{OA})}{1 + 2}$$vecOC = frac1(2hati+4hatj+4hatk) + 2(2hati+2hatj+hatk)3$$\vec{OC} = \frac{1(2\hat{i}+4\hat{j}+4\hat{k}) + 2(2\hat{i}+2\hat{j}+\hat{k})}{3}$$vecOC = frac(2+4)hati + (4+4)hatj + (4+2)hatk3$$\vec{OC} = \frac{(2+4)\hat{i} + (4+4)\hat{j} + (4+2)\hat{k}}{3}$$vecOC = frac6hati + 8hatj + 6hatk3 = 2hati + frac83hatj + 2hatk$$\vec{OC} = \frac{6\hat{i} + 8\hat{j} + 6\hat{k}}{3} = 2\hat{i} + \frac{8}{3}\hat{j} + 2\hat{k}$$
### Step 2: Compute Length of OC
Now, find the magnitude (length) of the vector vecOC$\vec{OC}$:
|vecOC| = sqrt2^2 + left(frac83right)^2 + 2^2$$|\vec{OC}| = \sqrt{2^2 + \left(\frac{8}{3}\right)^2 + 2^2}$$= sqrt4 + frac649 + 4 = sqrt8 + frac649$$= \sqrt{4 + \frac{64}{9} + 4} = \sqrt{8 + \frac{64}{9}}$$= sqrtfrac72 + 649 = sqrtfrac1369$$= \sqrt{\frac{72 + 64}{9}} = \sqrt{\frac{136}{9}}$$= fracsqrt4 times 343 = frac2sqrt343$$= \frac{\sqrt{4 \times 34}}{3} = \frac{2\sqrt{34}}{3}$$
### Pattern Recognition
Vector angle bisector questions invariably test the geometric property that the bisector divides the opposite side in the ratio of the side lengths. Combine this directly with the 3D coordinate section formula.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Vector Algebra
Q4jee_main_2024_30_january_eveningCross Product
Let veca = hati +alpha hatj +beta hatk,alpha ,beta in mathbbR$\vec{a} = \hat{i} +\alpha \hat{j} +\beta \hat{k},\alpha ,\beta \in \mathbb{R}$ .Let a vector vecb$\vec{b}$ be such that the angle between veca$\vec{a}$ and vecb$\vec{b}$ is fracpi4$\frac{\pi}{4}$ and |vecb |^2 = 6$|\vec{b} |^2 = 6$ If vecacdotvecb = 3sqrt2$\vec{a}\cdot\vec{b} = 3\sqrt{2}$ , then the value of (alpha^2 + beta^2)|vecatimes vecb |^2$(\alpha^{2} + \beta^{2})|\vec{a}\times \vec{b} |^{2}$ is equal to
Let veca$\vec{a}$ and vecb$\vec{b}$ be two vectors such that |vecb| = 1$|\vec{b}| = 1$ and |vecb times veca| = 2$|\vec{b} \times \vec{a}| = 2$ . Then left|left(vecb times vecaright) - vecbright|^2$\left|\left(\vec{b} \times \vec{a}\right) - \vec{b}\right|^2$ is equal to
A.3$3$
B.5$5$
C.1$1$
D.4$4$
Solution
### Related Formula
|vecx - vecy|^2 = |vecx|^2 + |vecy|^2 - 2(vecx cdot vecy)$$|\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2(\vec{x} \cdot \vec{y})$$(vecb times veca) cdot vecb = 0 quad text(Scalar triple product with repeated vectors is zero)$$(\vec{b} \times \vec{a}) \cdot \vec{b} = 0 \quad \text{(Scalar triple product with repeated vectors is zero)}$$
### Core Logic
Given |vecb| = 1$|\vec{b}| = 1$ and |vecb times veca| = 2$|\vec{b} \times \vec{a}| = 2$.
Expand the requested expression:
|(vecb times veca) - vecb|^2 = |vecb times veca|^2 + |vecb|^2 - 2((vecb times veca) cdot vecb)$$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = |\vec{b} \times \vec{a}|^2 + |\vec{b}|^2 - 2((\vec{b} \times \vec{a}) \cdot \vec{b})$$
### Step 1: Simplify using Vector Properties
The cross product (vecb times veca)$(\vec{b} \times \vec{a})$ produces a vector orthogonal to both vecb$\vec{b}$ and veca$\vec{a}$.
Therefore, (vecb times veca) cdot vecb = 0$(\vec{b} \times \vec{a}) \cdot \vec{b} = 0$.
Substituting this back:
|(vecb times veca) - vecb|^2 = |vecb times veca|^2 + |vecb|^2 - 0$$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = |\vec{b} \times \vec{a}|^2 + |\vec{b}|^2 - 0$$|(vecb times veca) - vecb|^2 = (2)^2 + (1)^2$$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = (2)^2 + (1)^2$$|(vecb times veca) - vecb|^2 = 4 + 1 = 5$$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = 4 + 1 = 5$$
### Pattern Recognition
A cross product vecu times vecv$\vec{u} \times \vec{v}$ is inherently perpendicular to vecu$\vec{u}$. Any length squared involving (vecu times vecv) pm vecu$(\vec{u} \times \vec{v}) \pm \vec{u}$ resolves simply via Pythagorean sum.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Vector Algebra
Let veca = a_1hati + a_2hatj + a_3hatk$\vec{a} = a_{1}\hat{i} + a_{2}\hat{j} + a_{3}\hat{k}$ and vecb = b_1hati + b_2hatj + b_3hatk$\vec{b} = b_{1}\hat{i} + b_{2}\hat{j} + b_{3}\hat{k}$ be two vectors such that |veca| = 1$|\vec{a}| = 1$; vecacdotvecb = 2$\vec{a}\cdot\vec{b} = 2$ and |vecb| = 4$|\vec{b}| = 4$. If vecc = 2(veca times vecb) - 3vecb$\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}$, then the angle between vecb$\vec{b}$ and vecc$\vec{c}$ is equal to:
### Related Formula
|veca times vecb|^2 + (veca cdot vecb)^2 = |veca|^2 |vecb|^2$$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$$cos theta = fracvecb cdot vecc|vecb| |vecc|$$\cos \theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|}$$
### Core Logic
Given |veca| = 1, |vecb| = 4, veca cdot vecb = 2$|\vec{a}| = 1, |\vec{b}| = 4, \vec{a} \cdot \vec{b} = 2$.
Let's evaluate |veca times vecb|^2$|\vec{a} \times \vec{b}|^2$ using Lagrange's identity:
|veca times vecb|^2 = |veca|^2 |vecb|^2 - (veca cdot vecb)^2$$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$$|veca times vecb|^2 = (1)(16) - (2)^2 = 16 - 4 = 12$$|\vec{a} \times \vec{b}|^2 = (1)(16) - (2)^2 = 16 - 4 = 12$$
### Step 1: Dot products with c
We are given:
vecc = 2(veca times vecb) - 3vecb$$\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}$$
To find the angle between vecb$\vec{b}$ and vecc$\vec{c}$, we need vecb cdot vecc$\vec{b} \cdot \vec{c}$ and |vecc|$|\vec{c}|$.
Taking the dot product with vecb$\vec{b}$ on both sides:
vecb cdot vecc = 2(vecb cdot (veca times vecb)) - 3(vecb cdot vecb)$$\vec{b} \cdot \vec{c} = 2(\vec{b} \cdot (\vec{a} \times \vec{b})) - 3(\vec{b} \cdot \vec{b})$$
Since vecb cdot (veca times vecb) = 0$\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$ (scalar triple product with repeated vector):
vecb cdot vecc = 0 - 3|vecb|^2 = -3(16) = -48 quad dots (1)$$\vec{b} \cdot \vec{c} = 0 - 3|\vec{b}|^2 = -3(16) = -48 \quad \dots (1)$$
### Step 2: Finding magnitude of c
Now, let's find |vecc|^2$|\vec{c}|^2$:
|vecc|^2 = vecc cdot vecc = (2(veca times vecb) - 3vecb) cdot (2(veca times vecb) - 3vecb)$$|\vec{c}|^2 = \vec{c} \cdot \vec{c} = (2(\vec{a} \times \vec{b}) - 3\vec{b}) \cdot (2(\vec{a} \times \vec{b}) - 3\vec{b})$$
Since (veca times vecb) cdot vecb = 0$(\vec{a} \times \vec{b}) \cdot \vec{b} = 0$, the cross terms vanish:
|vecc|^2 = 4|veca times vecb|^2 + 9|vecb|^2$$|\vec{c}|^2 = 4|\vec{a} \times \vec{b}|^2 + 9|\vec{b}|^2$$|vecc|^2 = 4(12) + 9(16) = 48 + 144 = 192$$|\vec{c}|^2 = 4(12) + 9(16) = 48 + 144 = 192$$|vecc| = sqrt192 = 8sqrt3$$|\vec{c}| = \sqrt{192} = 8\sqrt{3}$$
### Step 3: Calculating angle
Now apply the angle formula:
cos theta = fracvecb cdot vecc|vecb| |vecc| = frac-48(4)(8sqrt3)$$\cos \theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|} = \frac{-48}{(4)(8\sqrt{3})}$$cos theta = frac-4832sqrt3 = frac-32sqrt3 = -fracsqrt32$$\cos \theta = \frac{-48}{32\sqrt{3}} = \frac{-3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$$theta = cos^-1left(-fracsqrt32right)$$\theta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$$
### Pattern Recognition
Always remember that the cross product vector (veca times vecb)$(\vec{a} \times \vec{b})$ is orthogonal to both veca$\vec{a}$ and vecb$\vec{b}$. This immediately eliminates cross terms when finding the magnitude of linear combinations like x(veca times vecb) + yvecb$x(\vec{a} \times \vec{b}) + y\vec{b}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Vector Algebra
Q18jee_main_2024_30_jan_morningCross Product
Let A (2, 3, 5)$A (2, 3, 5)$ and C(-3, 4, -2)$C(-3, 4, -2)$ be opposite vertices of a parallelogram ABCD if the diagonal overrightarrowBD = hati + 2hatj + 3hatk$\overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$ then the area of the parallelogram is equal to
A.frac12sqrt410$\frac{1}{2}\sqrt{410}$
B.frac12sqrt474$\frac{1}{2}\sqrt{474}$
C.frac12sqrt586$\frac{1}{2}\sqrt{586}$
D.frac12sqrt306$\frac{1}{2}\sqrt{306}$
Solution
### Related Formula
textArea of parallelogram = frac12 |overrightarrowd_1 times overrightarrowd_2|$$\text{Area of parallelogram} = \frac{1}{2} |\overrightarrow{d_1} \times \overrightarrow{d_2}|$$
### Core Logic
The diagonals of the parallelogram are overrightarrowAC$\overrightarrow{AC}$ and overrightarrowBD$\overrightarrow{BD}$.
First, calculate the diagonal vector overrightarrowAC$\overrightarrow{AC}$:
overrightarrowAC = textPosition vector of C - textPosition vector of A$$\overrightarrow{AC} = \text{Position vector of } C - \text{Position vector of } A$$overrightarrowAC = (-3 - 2)hati + (4 - 3)hatj + (-2 - 5)hatk = -5hati + 1hatj - 7hatk$$\overrightarrow{AC} = (-3 - 2)\hat{i} + (4 - 3)\hat{j} + (-2 - 5)\hat{k} = -5\hat{i} + 1\hat{j} - 7\hat{k}$$
Alternatively, taking overrightarrowCA = 5hati - hatj + 7hatk$\overrightarrow{CA} = 5\hat{i} - \hat{j} + 7\hat{k}$. Let's use overrightarrowCA$\overrightarrow{CA}$ or overrightarrowAC$\overrightarrow{AC}$, the magnitude of the cross product will be the same.
The second diagonal is given:
overrightarrowBD = hati + 2hatj + 3hatk$$\overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$$
### Step 1: Finding Cross Product
overrightarrowArea = frac12 |overrightarrowAC times overrightarrowBD|$$\overrightarrow{Area} = \frac{1}{2} |\overrightarrow{AC} \times \overrightarrow{BD}|$$overrightarrowAC times overrightarrowBD = beginvmatrix hati & hatj & hatk \\ -5 & 1 & -7 \\ 1 & 2 & 3 endvmatrix$$\overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -7 \\ 1 & 2 & 3 \end{vmatrix}$$= hati(3 - (-14)) - hatj(-15 - (-7)) + hatk(-10 - 1)$$= \hat{i}(3 - (-14)) - \hat{j}(-15 - (-7)) + \hat{k}(-10 - 1)$$= hati(17) - hatj(-8) + hatk(-11)$$= \hat{i}(17) - \hat{j}(-8) + \hat{k}(-11)$$= 17hati + 8hatj - 11hatk$$= 17\hat{i} + 8\hat{j} - 11\hat{k}$$
### Step 2: Calculating Magnitude
textArea = frac12 |17hati + 8hatj - 11hatk|$$\text{Area} = \frac{1}{2} |17\hat{i} + 8\hat{j} - 11\hat{k}|$$= frac12 sqrt17^2 + 8^2 + (-11)^2$$= \frac{1}{2} \sqrt{17^2 + 8^2 + (-11)^2}$$= frac12 sqrt289 + 64 + 121$$= \frac{1}{2} \sqrt{289 + 64 + 121}$$= frac12 sqrt474$$= \frac{1}{2} \sqrt{474}$$
### Pattern Recognition
When opposite vertices and one full diagonal vector are provided, immediately calculate the second diagonal vector via displacement and evaluate half the magnitude of their cross product.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Vector Algebra
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