Math / Geometry

Question13 Diberi segi tiga $A B C$ dengan $\overrightarrow{A B}=4 \underset{\sim}{i}-6 \underset{\sim}{j}$ dan $\overrightarrow{A C}=2 \underset{\sim}{i}+4 \underset{\sim}{j} . T$ berada pada garis $B C$ dengan keadaan $3 B T=T C$ . It is given that a triangle $A B C$ with $\overrightarrow{A B}=4 \underset{\sim}{i}-6 \underset{\sim}{j} \underset{\sim n d}{A C}=2 \underset{\sim}{i}+4 \underset{\sim}{j}$ . T lies on the line $B C$ such that $3 B T=T C$ . (a) Cari vektor Find the vector (i) $\overrightarrow{B C}$ , (ii) $\overrightarrow{A T}$ . Seterusnya, car1 vektor unit dalam arah $\overrightarrow{A T}$ . $\overrightarrow{A T}$ . Hence, find the unit vector in the direction of $\overrightarrow{A T}$ . [5 markah] [5 marks] (b) Jika $D$ ialah satu titik dengan keadaan $\overrightarrow{A D}=h \overrightarrow{B C}$ dan $\overrightarrow{A D}=-3 \underset{\sim}{i}+\underset{\sim}{k j}$ , dengan keadaan $h$ dan $k$ adalah pemalar. Cari nilai $h$ dan nilai $k$ . If $D$ is a point such that $\overrightarrow{A D}=h \overrightarrow{B C}$ and $\overrightarrow{A D}=-3 i \underset{\sim}{i}+k$ , such that $h$ and $k$ are constants. Find the value of $h$ and of $k$ .

Studdy Solution

STEP 1

What is this asking? We're given two sides of a triangle as vectors, and a point $T$ on the third side with a specific ratio.
We need to find some other vectors and a unit vector.
Then, we're given another point $D$ and its vector related to the third side of the triangle, and we need to find some constants. Watch out! Vector addition and subtraction can be tricky!
Remember, it's tip minus tail.
Also, don't forget to normalize your unit vector!

STEP 2

1. Find BC
2. Find AT and its unit vector
3. Find h and k

STEP 3

Alright, let's start by finding $\overrightarrow{BC}$ !
We know that $\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC}$ .
Remember, it's tip minus tail, so $\overrightarrow{BA}$ is the opposite of $\overrightarrow{AB}$ .

STEP 4

So, $\overrightarrow{BA} = -\overrightarrow{AB} = -(4\mathbf{i} - 6\mathbf{j}) = -4\mathbf{i} + 6\mathbf{j}$ .

STEP 5

Now, add the vectors: $\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC} = (-4\mathbf{i} + 6\mathbf{j}) + (2\mathbf{i} + 4\mathbf{j}) = (-4+2)\mathbf{i} + (6+4)\mathbf{j} = -2\mathbf{i} + 10\mathbf{j}$ .
So, $\overrightarrow{BC} = -2\mathbf{i} + 10\mathbf{j}$ !

STEP 6

We're given that $3\overrightarrow{BT} = \overrightarrow{TC}$ .
This means $\overrightarrow{BT}$ is one-fourth of $\overrightarrow{BC}$ since the segments are in a 1:3 ratio.
So, $\overrightarrow{BT} = \frac{1}{4}\overrightarrow{BC} = \frac{1}{4}(-2\mathbf{i} + 10\mathbf{j}) = -\frac{1}{2}\mathbf{i} + \frac{5}{2}\mathbf{j}$ .

STEP 7

Now, let's find $\overrightarrow{AT}$ .
We know $\overrightarrow{AT} = \overrightarrow{AB} + \overrightarrow{BT}$ .
Substituting our known vectors: $\overrightarrow{AT} = (4\mathbf{i} - 6\mathbf{j}) + (-\frac{1}{2}\mathbf{i} + \frac{5}{2}\mathbf{j}) = (4-\frac{1}{2})\mathbf{i} + (-6+\frac{5}{2})\mathbf{j} = \frac{7}{2}\mathbf{i} - \frac{7}{2}\mathbf{j}$ .

STEP 8

To find the unit vector, we need to divide $\overrightarrow{AT}$ by its magnitude.
The magnitude of $\overrightarrow{AT}$ is $\sqrt{(\frac{7}{2})^2 + (-\frac{7}{2})^2} = \sqrt{\frac{49}{4} + \frac{49}{4}} = \sqrt{\frac{98}{4}} = \frac{7\sqrt{2}}{2}$ .

STEP 9

The unit vector in the direction of $\overrightarrow{AT}$ is $\frac{\overrightarrow{AT}}{|\overrightarrow{AT}|} = \frac{\frac{7}{2}\mathbf{i} - \frac{7}{2}\mathbf{j}}{\frac{7\sqrt{2}}{2}} = \frac{1}{\sqrt{2}}\mathbf{i} - \frac{1}{\sqrt{2}}\mathbf{j}$ .

STEP 10

We're given $\overrightarrow{AD} = h\overrightarrow{BC}$ and $\overrightarrow{AD} = -3\mathbf{i} + k\mathbf{j}$ .
We already found $\overrightarrow{BC} = -2\mathbf{i} + 10\mathbf{j}$ .

STEP 11

So, $-3\mathbf{i} + k\mathbf{j} = h(-2\mathbf{i} + 10\mathbf{j})$ .
This means $-3\mathbf{i} + k\mathbf{j} = -2h\mathbf{i} + 10h\mathbf{j}$ .

STEP 12

By comparing the coefficients of $\mathbf{i}$ , we have $-3 = -2h$ , so $h = \frac{3}{2}$ .

STEP 13

Comparing the coefficients of $\mathbf{j}$ , we have $k = 10h$ .
Since $h = \frac{3}{2}$ , we have $k = 10 \cdot \frac{3}{2} = 15$ .

STEP 14

(a)(i) $\overrightarrow{BC} = -2\mathbf{i} + 10\mathbf{j}$ (a)(ii) $\overrightarrow{AT} = \frac{7}{2}\mathbf{i} - \frac{7}{2}\mathbf{j}$ .
The unit vector is $\frac{1}{\sqrt{2}}\mathbf{i} - \frac{1}{\sqrt{2}}\mathbf{j}$ . (b) $h = \frac{3}{2}$ and $k = 15$ .

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Studdy Solution

STEP 1

STEP 2

1. Find BC
2. Find AT and its unit vector
3. Find h and k

STEP 3

Alright, let's start by finding $\overrightarrow{BC}$ !
We know that $\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC}$ .
Remember, it's tip minus tail, so $\overrightarrow{BA}$ is the opposite of $\overrightarrow{AB}$ .

STEP 4

So, $\overrightarrow{BA} = -\overrightarrow{AB} = -(4\mathbf{i} - 6\mathbf{j}) = -4\mathbf{i} + 6\mathbf{j}$ .

STEP 5

Now, add the vectors: $\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC} = (-4\mathbf{i} + 6\mathbf{j}) + (2\mathbf{i} + 4\mathbf{j}) = (-4+2)\mathbf{i} + (6+4)\mathbf{j} = -2\mathbf{i} + 10\mathbf{j}$ .
So, $\overrightarrow{BC} = -2\mathbf{i} + 10\mathbf{j}$ !

STEP 6

STEP 7

STEP 8

To find the unit vector, we need to divide $\overrightarrow{AT}$ by its magnitude.
The magnitude of $\overrightarrow{AT}$ is $\sqrt{(\frac{7}{2})^2 + (-\frac{7}{2})^2} = \sqrt{\frac{49}{4} + \frac{49}{4}} = \sqrt{\frac{98}{4}} = \frac{7\sqrt{2}}{2}$ .

STEP 9

The unit vector in the direction of $\overrightarrow{AT}$ is $\frac{\overrightarrow{AT}}{|\overrightarrow{AT}|} = \frac{\frac{7}{2}\mathbf{i} - \frac{7}{2}\mathbf{j}}{\frac{7\sqrt{2}}{2}} = \frac{1}{\sqrt{2}}\mathbf{i} - \frac{1}{\sqrt{2}}\mathbf{j}$ .

STEP 10

We're given $\overrightarrow{AD} = h\overrightarrow{BC}$ and $\overrightarrow{AD} = -3\mathbf{i} + k\mathbf{j}$ .
We already found $\overrightarrow{BC} = -2\mathbf{i} + 10\mathbf{j}$ .

STEP 11

So, $-3\mathbf{i} + k\mathbf{j} = h(-2\mathbf{i} + 10\mathbf{j})$ .
This means $-3\mathbf{i} + k\mathbf{j} = -2h\mathbf{i} + 10h\mathbf{j}$ .

STEP 12

By comparing the coefficients of $\mathbf{i}$ , we have $-3 = -2h$ , so $h = \frac{3}{2}$ .

STEP 13

Comparing the coefficients of $\mathbf{j}$ , we have $k = 10h$ .
Since $h = \frac{3}{2}$ , we have $k = 10 \cdot \frac{3}{2} = 15$ .

STEP 14

(a)(i) $\overrightarrow{BC} = -2\mathbf{i} + 10\mathbf{j}$ (a)(ii) $\overrightarrow{AT} = \frac{7}{2}\mathbf{i} - \frac{7}{2}\mathbf{j}$ .
The unit vector is $\frac{1}{\sqrt{2}}\mathbf{i} - \frac{1}{\sqrt{2}}\mathbf{j}$ . (b) $h = \frac{3}{2}$ and $k = 15$ .

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Studdy Solution

STEP 1

STEP 2

1. Find BC2. Find AT and its unit vector3. Find h and k

STEP 3

STEP 4

So, BAundefined=−ABundefined=−(4i−6j)=−4i+6j\overrightarrow{BA} = -\overrightarrow{AB} = -(4\mathbf{i} - 6\mathbf{j}) = -4\mathbf{i} + 6\mathbf{j}BA=−AB=−(4i−6j)=−4i+6j.

STEP 5

STEP 6

STEP 7

STEP 8

STEP 9

STEP 10

We're given ADundefined=hBCundefined\overrightarrow{AD} = h\overrightarrow{BC}AD=hBC and ADundefined=−3i+kj\overrightarrow{AD} = -3\mathbf{i} + k\mathbf{j}AD=−3i+kj.We already found BCundefined=−2i+10j\overrightarrow{BC} = -2\mathbf{i} + 10\mathbf{j}BC=−2i+10j.

STEP 11

So, −3i+kj=h(−2i+10j)-3\mathbf{i} + k\mathbf{j} = h(-2\mathbf{i} + 10\mathbf{j})−3i+kj=h(−2i+10j).This means −3i+kj=−2hi+10hj-3\mathbf{i} + k\mathbf{j} = -2h\mathbf{i} + 10h\mathbf{j}−3i+kj=−2hi+10hj.

STEP 12

By comparing the coefficients of i\mathbf{i}i, we have −3=−2h-3 = -2h−3=−2h, so h=32h = \frac{3}{2}h=23​.

STEP 13

Comparing the coefficients of j\mathbf{j}j, we have k=10hk = 10hk=10h.Since h=32h = \frac{3}{2}h=23​, we have k=10⋅32=15k = 10 \cdot \frac{3}{2} = 15k=10⋅23​=15.

STEP 14

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1. Find BC
2. Find AT and its unit vector
3. Find h and k

So, $\overrightarrow{BA} = -\overrightarrow{AB} = -(4\mathbf{i} - 6\mathbf{j}) = -4\mathbf{i} + 6\mathbf{j}$ .

We're given $\overrightarrow{AD} = h\overrightarrow{BC}$ and $\overrightarrow{AD} = -3\mathbf{i} + k\mathbf{j}$ .
We already found $\overrightarrow{BC} = -2\mathbf{i} + 10\mathbf{j}$ .

So, $-3\mathbf{i} + k\mathbf{j} = h(-2\mathbf{i} + 10\mathbf{j})$ .
This means $-3\mathbf{i} + k\mathbf{j} = -2h\mathbf{i} + 10h\mathbf{j}$ .

By comparing the coefficients of $\mathbf{i}$ , we have $-3 = -2h$ , so $h = \frac{3}{2}$ .

Comparing the coefficients of $\mathbf{j}$ , we have $k = 10h$ .
Since $h = \frac{3}{2}$ , we have $k = 10 \cdot \frac{3}{2} = 15$ .