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Question: If the position vectors of the vertices A, B and C of a \(\Delta ABC\) are respectively \(4\widehat{...

If the position vectors of the vertices A, B and C of a ΔABC\Delta ABC are respectively 4i^+7j^+8k^4\widehat{i}+7\widehat{j}+8\widehat{k}, 2i^+3j^+4k^2\widehat{i}+3\widehat{j}+4\widehat{k} and 2i^+5j^+7k^2\widehat{i}+5\widehat{j}+7\widehat{k}, then the position vector of the point, where the bisector of ∠A\angle A meets BC is :
A. 12(4i^+8j^+11k^)\dfrac{1}{2}\left( 4\widehat{i}+8\widehat{j}+11\widehat{k} \right)
B. 13(6i^+13j^+18k^)\dfrac{1}{3}\left( 6\widehat{i}+13\widehat{j}+18\widehat{k} \right)
C. 14(8i^+14j^+9k^)\dfrac{1}{4}\left( 8\widehat{i}+14\widehat{j}+9\widehat{k} \right)
D. 13(6i^+11j^+15k^)\dfrac{1}{3}\left( 6\widehat{i}+11\widehat{j}+15\widehat{k} \right)

Explanation

Solution

First, you need to draw the diagram and mention the given information in the diagram. Find the vectors for AB→\overrightarrow{AB} and AC→\overrightarrow{AC} and also find their magnitudes. Now, use the section formula of internal division method, r→ =mb→ +na→ m+n\overset{\to }{\mathop{r}}\,=\dfrac{m\overset{\to }{\mathop{b}}\,+n\overset{\to }{\mathop{a}}\,}{m+n}, substitute the obtained values and get the result.

Complete step-by-step solution:
Let us first draw the ΔABC\Delta ABC denoting the vectors A as 4i^+7j^+8k^4\widehat{i}+7\widehat{j}+8\widehat{k}, B as 2i^+3j^+4k^2\widehat{i}+3\widehat{j}+4\widehat{k} and C as 2i^+5j^+7k^2\widehat{i}+5\widehat{j}+7\widehat{k} and a bisector of ∠A\angle A which meets the side BC at a point D.

Since AD bisects the angle ∠A\angle A and divides BC in the ratio AB : BC. First, let us find the vectors AB→\overrightarrow{AB} and AC→\overrightarrow{AC},
AB→=(2i^+3j^+4k^)−(4i^+7j^+8k^) ⇒AB→=(2i^+3j^+4k^)−4i^−7j^−8k^ ⇒AB→=−2i^−4j^−4k^ \begin{aligned} & \overrightarrow{AB}=\left( 2\widehat{i}+3\widehat{j}+4\widehat{k} \right)-\left( 4\widehat{i}+7\widehat{j}+8\widehat{k} \right) \\\ & \Rightarrow \overrightarrow{AB}=\left( 2\widehat{i}+3\widehat{j}+4\widehat{k} \right)-4\widehat{i}-7\widehat{j}-8\widehat{k} \\\ & \Rightarrow \overrightarrow{AB}=-2\widehat{i}-4\widehat{j}-4\widehat{k} \\\ \end{aligned}
Similarly, let us find the vector AC→\overrightarrow{AC},
AC→=(2i^+5j^+7k^)−(4i^+7j^+8k^) ⇒AC→=2i^+5j^+7k^−4i^−7j^−8k^ ⇒AC→=−2i^−2j^−k^ \begin{aligned} & \overrightarrow{AC}=\left( 2\widehat{i}+5\widehat{j}+7\widehat{k} \right)-\left( 4\widehat{i}+7\widehat{j}+8\widehat{k} \right) \\\ & \Rightarrow \overrightarrow{AC}=2\widehat{i}+5\widehat{j}+7\widehat{k}-4\widehat{i}-7\widehat{j}-8\widehat{k} \\\ & \Rightarrow \overrightarrow{AC}=-2\widehat{i}-2\widehat{j}-\widehat{k} \\\ \end{aligned}
Now, let us find the magnitude of the vectors AB→\overrightarrow{AB} and AC→\overrightarrow{AC},
∣AB→∣=(−2)2+(−4)2+(−4)2 ⇒∣AB→∣=4+16+16 ⇒∣AB→∣=36 ⇒∣AB→∣=6 \begin{aligned} & \left| \overrightarrow{AB} \right|=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -4 \right)}^{2}}+{{\left( -4 \right)}^{2}}} \\\ & \Rightarrow \left| \overrightarrow{AB} \right|=\sqrt{4+16+16} \\\ & \Rightarrow \left| \overrightarrow{AB} \right|=\sqrt{36} \\\ & \Rightarrow \left| \overrightarrow{AB} \right|=6 \\\ \end{aligned}
Similarly, the magnitude of vector AC→\overrightarrow{AC} will be,
∣AC→∣=(−2)2+(−2)2+(−1)2 ⇒∣AC→∣=4+4+1 ⇒∣AC→∣=9 ⇒∣AC→∣=3 \begin{aligned} & \left| \overrightarrow{AC} \right|=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -2 \right)}^{2}}+{{\left( -1 \right)}^{2}}} \\\ & \Rightarrow \left| \overrightarrow{AC} \right|=\sqrt{4+4+1} \\\ & \Rightarrow \left| \overrightarrow{AC} \right|=\sqrt{9} \\\ & \Rightarrow \left| \overrightarrow{AC} \right|=3 \\\ \end{aligned}
By using section formula of internal division method, we know,
r→ =mb→ +na→ m+n………(i)\overset{\to }{\mathop{r}}\,=\dfrac{m\overset{\to }{\mathop{b}}\,+n\overset{\to }{\mathop{a}}\,}{m+n}\ldots \ldots \ldots \left( i \right)
Here, according to the above expression, we have,
m=∣AB↔∣=6 n=∣AC↔∣=3 \begin{aligned} & m=\left| \overleftrightarrow{AB} \right|=6 \\\ & n=\left| \overleftrightarrow{AC} \right|=3 \\\ \end{aligned}
Let us consider a→ \overset{\to }{\mathop{a}}\, as B→ \overset{\to }{\mathop{B}}\, and b→ \overset{\to }{\mathop{b}}\, as C→ \overset{\to }{\mathop{C}}\, and r→ \overset{\to }{\mathop{r}}\, as D→ \overset{\to }{\mathop{D}}\,. Therefore, we get,
a→ =2i^+3j^+4k^\overset{\to }{\mathop{a}}\,=2\widehat{i}+3\widehat{j}+4\widehat{k} and b→ =2i^+5j^+7k^\overset{\to }{\mathop{b}}\,=2\widehat{i}+5\widehat{j}+7\widehat{k}
Now, when we substitute the values of m and n, the vectors a→ \overset{\to }{\mathop{a}}\, and b→ \overset{\to }{\mathop{b}}\, in the equation (i), we will get the position vector of D, which is,
D→ =6(2i^+5j^+7k^)+3(2i^+3j^+4k^)6+3 ⇒D→ =3[2(2i^+5j^+7k^)+(2i^+3j^+4k^)]9 ⇒D→ =13[4i^+10j^+14k^+2i^+3j^+4k^] ⇒D→ =13[6i^+13j^+18k^] \begin{aligned} & \overset{\to }{\mathop{D}}\,=\dfrac{6\left( 2\widehat{i}+5\widehat{j}+7\widehat{k} \right)+3\left( 2\widehat{i}+3\widehat{j}+4\widehat{k} \right)}{6+3} \\\ & \Rightarrow \overset{\to }{\mathop{D}}\,=\dfrac{3\left[ 2\left( 2\widehat{i}+5\widehat{j}+7\widehat{k} \right)+\left( 2\widehat{i}+3\widehat{j}+4\widehat{k} \right) \right]}{9} \\\ & \Rightarrow \overset{\to }{\mathop{D}}\,=\dfrac{1}{3}\left[ 4\widehat{i}+10\widehat{j}+14\widehat{k}+2\widehat{i}+3\widehat{j}+4\widehat{k} \right] \\\ & \Rightarrow \overset{\to }{\mathop{D}}\,=\dfrac{1}{3}\left[ 6\widehat{i}+13\widehat{j}+18\widehat{k} \right] \\\ \end{aligned}
Therefore, the position vector of D is 13[6i^+13j^+18k^]\dfrac{1}{3}\left[ 6\widehat{i}+13\widehat{j}+18\widehat{k} \right].
Hence, the correct option is option B.

Note: In this question, always draw the diagram which makes it easy to understand the question. You also need to know, how to get the position vectors and how to find the magnitude of the vectors. The vector in the form of I, j, k represents the position of a vector in three-dimensional graph y x, y, z respectively.