87991
If \(\hat{i}+\hat{j}, \hat{j}+\hat{k}, \hat{i}+\hat{k}\) are the position vectors of the vertices of a triangle \(A B C\) taken in order, then \(\angle \mathrm{A}\) is equal to
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{5}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{3}\)
Explanation:
(D) : Given position vector of the vertices, \(\overrightarrow{\mathrm{OA}}=\hat{i}+\hat{j} \Rightarrow \overrightarrow{\mathrm{OB}}=\hat{j}+\hat{k} \Rightarrow \overrightarrow{\mathrm{OC}}=\hat{i}+\hat{k}\) \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(\hat{j}+\hat{k})-(\hat{i}+\hat{j})\) \(=\hat{k}-\hat{i}\) \(\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=(\hat{\mathrm{i}}+\hat{\mathrm{k}})-(\hat{\mathrm{i}}+\hat{\mathrm{j}})\) \(=\hat{\mathrm{k}}-\hat{\mathrm{j}}\) We know that, \(\cos \theta=\frac{\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{AB}}}{|\overrightarrow{\mathrm{AC}}| \cdot|\overrightarrow{\mathrm{AB}}|}=\frac{(\hat{\mathrm{k}}-\hat{\mathrm{j}}) \cdot(\hat{\mathrm{k}}-\hat{\mathrm{i}})}{|(\hat{\mathrm{k}}-\hat{\mathrm{j}})||(\hat{\mathrm{k}}-\hat{\mathrm{i}})|}\) \(=\frac{\hat{\mathrm{k}} \hat{\mathrm{k}}-\hat{\mathrm{k}} \cdot \hat{\mathrm{i}}-\hat{\mathrm{j}} \hat{\mathrm{k}}+\hat{\mathrm{j}} \cdot \hat{\mathrm{i}}}{\sqrt{(1)^2+(-1)^2} \sqrt{(1)^2+(-1)^2}} \quad \because(\hat{\mathrm{k}} \cdot \hat{\mathrm{k}}=1)\) \(\cos \theta=\frac{1-0-0+0}{\sqrt{2} \sqrt{2}}=\frac{1}{2}\) \(\cos \theta=60^{\circ} \text { or } \theta=\frac{\pi}{3}\)
BITSAT-2018
Vector Algebra
87992
Let \(a, b\) and \(c\) be three vectors satisfying \(a \times b\) \(=(a \times c),|a|=|c|=1,|b|=4\) and \(|b \times c|=\sqrt{15}\). If \(b-2 c=\lambda a\), then \(\lambda\) equals
1 1
2 -1
3 2
4 -4
Explanation:
(D)Given, \(\mathrm{a} \times \mathrm{b}=(\mathrm{a} \times \mathrm{c}),\) \(|\mathrm{a}|=|\mathrm{c}|=1,\) \(|\mathrm{~b}|=4\) \(|\mathrm{~b} \times \mathrm{c}|=\sqrt{15}\) \(\mathrm{~b}-2 \mathrm{c}=\lambda \mathrm{a}\) \(\lambda=?\) Let \(\theta\) be the angle between \(b\) and \(c\). \(\because \quad|\mathrm{b} \times \mathrm{c}|=\sqrt{15}\) \(|\mathrm{~b}||\mathrm{c}| \sin \theta=\sqrt{15}\) \(\sin \theta=\frac{\sqrt{15}}{|\mathrm{~b}||\mathrm{c}|}\) \(\sin \theta=\frac{\sqrt{15}}{4 \times 1}=\frac{\sqrt{15}}{4}\) \(\therefore \cos \theta=\sqrt{1-\frac{15}{16}}=\frac{1}{\sqrt{16}}=\frac{1}{4}\) \(\mathrm{~b}-2 \mathrm{c}=\lambda \mathrm{a}\) (Given) Squaring in both sides, \(|\mathrm{b}-2 \mathrm{c}|^2=|\lambda \mathrm{a}|^2\) \(|\mathrm{~b}|^2+4|\mathrm{c}|^2-4(\mathrm{~b} \cdot \mathrm{c})=\lambda^2|\mathrm{a}|^2\) \(16+4-4|\mathrm{~b}||\mathrm{c}| \cos \theta=\lambda^2\) \(20-16 \cos \theta=\lambda^2\) \(20-16 \times \frac{1}{4}=\lambda^2\) \(20-4=\lambda^2\) \(\lambda^2=16\) \(\lambda= \pm 4\)
BITSAT-2015
Vector Algebra
87993
If \(|\vec{a}|=3,|\vec{b}|=4\), then a value of \(\lambda\), for which \(\vec{a}+\lambda \vec{b}\) is perpendicular to \(\vec{a}-\lambda \vec{b}\)
87991
If \(\hat{i}+\hat{j}, \hat{j}+\hat{k}, \hat{i}+\hat{k}\) are the position vectors of the vertices of a triangle \(A B C\) taken in order, then \(\angle \mathrm{A}\) is equal to
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{5}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{3}\)
Explanation:
(D) : Given position vector of the vertices, \(\overrightarrow{\mathrm{OA}}=\hat{i}+\hat{j} \Rightarrow \overrightarrow{\mathrm{OB}}=\hat{j}+\hat{k} \Rightarrow \overrightarrow{\mathrm{OC}}=\hat{i}+\hat{k}\) \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(\hat{j}+\hat{k})-(\hat{i}+\hat{j})\) \(=\hat{k}-\hat{i}\) \(\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=(\hat{\mathrm{i}}+\hat{\mathrm{k}})-(\hat{\mathrm{i}}+\hat{\mathrm{j}})\) \(=\hat{\mathrm{k}}-\hat{\mathrm{j}}\) We know that, \(\cos \theta=\frac{\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{AB}}}{|\overrightarrow{\mathrm{AC}}| \cdot|\overrightarrow{\mathrm{AB}}|}=\frac{(\hat{\mathrm{k}}-\hat{\mathrm{j}}) \cdot(\hat{\mathrm{k}}-\hat{\mathrm{i}})}{|(\hat{\mathrm{k}}-\hat{\mathrm{j}})||(\hat{\mathrm{k}}-\hat{\mathrm{i}})|}\) \(=\frac{\hat{\mathrm{k}} \hat{\mathrm{k}}-\hat{\mathrm{k}} \cdot \hat{\mathrm{i}}-\hat{\mathrm{j}} \hat{\mathrm{k}}+\hat{\mathrm{j}} \cdot \hat{\mathrm{i}}}{\sqrt{(1)^2+(-1)^2} \sqrt{(1)^2+(-1)^2}} \quad \because(\hat{\mathrm{k}} \cdot \hat{\mathrm{k}}=1)\) \(\cos \theta=\frac{1-0-0+0}{\sqrt{2} \sqrt{2}}=\frac{1}{2}\) \(\cos \theta=60^{\circ} \text { or } \theta=\frac{\pi}{3}\)
BITSAT-2018
Vector Algebra
87992
Let \(a, b\) and \(c\) be three vectors satisfying \(a \times b\) \(=(a \times c),|a|=|c|=1,|b|=4\) and \(|b \times c|=\sqrt{15}\). If \(b-2 c=\lambda a\), then \(\lambda\) equals
1 1
2 -1
3 2
4 -4
Explanation:
(D)Given, \(\mathrm{a} \times \mathrm{b}=(\mathrm{a} \times \mathrm{c}),\) \(|\mathrm{a}|=|\mathrm{c}|=1,\) \(|\mathrm{~b}|=4\) \(|\mathrm{~b} \times \mathrm{c}|=\sqrt{15}\) \(\mathrm{~b}-2 \mathrm{c}=\lambda \mathrm{a}\) \(\lambda=?\) Let \(\theta\) be the angle between \(b\) and \(c\). \(\because \quad|\mathrm{b} \times \mathrm{c}|=\sqrt{15}\) \(|\mathrm{~b}||\mathrm{c}| \sin \theta=\sqrt{15}\) \(\sin \theta=\frac{\sqrt{15}}{|\mathrm{~b}||\mathrm{c}|}\) \(\sin \theta=\frac{\sqrt{15}}{4 \times 1}=\frac{\sqrt{15}}{4}\) \(\therefore \cos \theta=\sqrt{1-\frac{15}{16}}=\frac{1}{\sqrt{16}}=\frac{1}{4}\) \(\mathrm{~b}-2 \mathrm{c}=\lambda \mathrm{a}\) (Given) Squaring in both sides, \(|\mathrm{b}-2 \mathrm{c}|^2=|\lambda \mathrm{a}|^2\) \(|\mathrm{~b}|^2+4|\mathrm{c}|^2-4(\mathrm{~b} \cdot \mathrm{c})=\lambda^2|\mathrm{a}|^2\) \(16+4-4|\mathrm{~b}||\mathrm{c}| \cos \theta=\lambda^2\) \(20-16 \cos \theta=\lambda^2\) \(20-16 \times \frac{1}{4}=\lambda^2\) \(20-4=\lambda^2\) \(\lambda^2=16\) \(\lambda= \pm 4\)
BITSAT-2015
Vector Algebra
87993
If \(|\vec{a}|=3,|\vec{b}|=4\), then a value of \(\lambda\), for which \(\vec{a}+\lambda \vec{b}\) is perpendicular to \(\vec{a}-\lambda \vec{b}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Vector Algebra
87991
If \(\hat{i}+\hat{j}, \hat{j}+\hat{k}, \hat{i}+\hat{k}\) are the position vectors of the vertices of a triangle \(A B C\) taken in order, then \(\angle \mathrm{A}\) is equal to
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{5}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{3}\)
Explanation:
(D) : Given position vector of the vertices, \(\overrightarrow{\mathrm{OA}}=\hat{i}+\hat{j} \Rightarrow \overrightarrow{\mathrm{OB}}=\hat{j}+\hat{k} \Rightarrow \overrightarrow{\mathrm{OC}}=\hat{i}+\hat{k}\) \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(\hat{j}+\hat{k})-(\hat{i}+\hat{j})\) \(=\hat{k}-\hat{i}\) \(\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=(\hat{\mathrm{i}}+\hat{\mathrm{k}})-(\hat{\mathrm{i}}+\hat{\mathrm{j}})\) \(=\hat{\mathrm{k}}-\hat{\mathrm{j}}\) We know that, \(\cos \theta=\frac{\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{AB}}}{|\overrightarrow{\mathrm{AC}}| \cdot|\overrightarrow{\mathrm{AB}}|}=\frac{(\hat{\mathrm{k}}-\hat{\mathrm{j}}) \cdot(\hat{\mathrm{k}}-\hat{\mathrm{i}})}{|(\hat{\mathrm{k}}-\hat{\mathrm{j}})||(\hat{\mathrm{k}}-\hat{\mathrm{i}})|}\) \(=\frac{\hat{\mathrm{k}} \hat{\mathrm{k}}-\hat{\mathrm{k}} \cdot \hat{\mathrm{i}}-\hat{\mathrm{j}} \hat{\mathrm{k}}+\hat{\mathrm{j}} \cdot \hat{\mathrm{i}}}{\sqrt{(1)^2+(-1)^2} \sqrt{(1)^2+(-1)^2}} \quad \because(\hat{\mathrm{k}} \cdot \hat{\mathrm{k}}=1)\) \(\cos \theta=\frac{1-0-0+0}{\sqrt{2} \sqrt{2}}=\frac{1}{2}\) \(\cos \theta=60^{\circ} \text { or } \theta=\frac{\pi}{3}\)
BITSAT-2018
Vector Algebra
87992
Let \(a, b\) and \(c\) be three vectors satisfying \(a \times b\) \(=(a \times c),|a|=|c|=1,|b|=4\) and \(|b \times c|=\sqrt{15}\). If \(b-2 c=\lambda a\), then \(\lambda\) equals
1 1
2 -1
3 2
4 -4
Explanation:
(D)Given, \(\mathrm{a} \times \mathrm{b}=(\mathrm{a} \times \mathrm{c}),\) \(|\mathrm{a}|=|\mathrm{c}|=1,\) \(|\mathrm{~b}|=4\) \(|\mathrm{~b} \times \mathrm{c}|=\sqrt{15}\) \(\mathrm{~b}-2 \mathrm{c}=\lambda \mathrm{a}\) \(\lambda=?\) Let \(\theta\) be the angle between \(b\) and \(c\). \(\because \quad|\mathrm{b} \times \mathrm{c}|=\sqrt{15}\) \(|\mathrm{~b}||\mathrm{c}| \sin \theta=\sqrt{15}\) \(\sin \theta=\frac{\sqrt{15}}{|\mathrm{~b}||\mathrm{c}|}\) \(\sin \theta=\frac{\sqrt{15}}{4 \times 1}=\frac{\sqrt{15}}{4}\) \(\therefore \cos \theta=\sqrt{1-\frac{15}{16}}=\frac{1}{\sqrt{16}}=\frac{1}{4}\) \(\mathrm{~b}-2 \mathrm{c}=\lambda \mathrm{a}\) (Given) Squaring in both sides, \(|\mathrm{b}-2 \mathrm{c}|^2=|\lambda \mathrm{a}|^2\) \(|\mathrm{~b}|^2+4|\mathrm{c}|^2-4(\mathrm{~b} \cdot \mathrm{c})=\lambda^2|\mathrm{a}|^2\) \(16+4-4|\mathrm{~b}||\mathrm{c}| \cos \theta=\lambda^2\) \(20-16 \cos \theta=\lambda^2\) \(20-16 \times \frac{1}{4}=\lambda^2\) \(20-4=\lambda^2\) \(\lambda^2=16\) \(\lambda= \pm 4\)
BITSAT-2015
Vector Algebra
87993
If \(|\vec{a}|=3,|\vec{b}|=4\), then a value of \(\lambda\), for which \(\vec{a}+\lambda \vec{b}\) is perpendicular to \(\vec{a}-\lambda \vec{b}\)
87991
If \(\hat{i}+\hat{j}, \hat{j}+\hat{k}, \hat{i}+\hat{k}\) are the position vectors of the vertices of a triangle \(A B C\) taken in order, then \(\angle \mathrm{A}\) is equal to
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{5}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{3}\)
Explanation:
(D) : Given position vector of the vertices, \(\overrightarrow{\mathrm{OA}}=\hat{i}+\hat{j} \Rightarrow \overrightarrow{\mathrm{OB}}=\hat{j}+\hat{k} \Rightarrow \overrightarrow{\mathrm{OC}}=\hat{i}+\hat{k}\) \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(\hat{j}+\hat{k})-(\hat{i}+\hat{j})\) \(=\hat{k}-\hat{i}\) \(\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=(\hat{\mathrm{i}}+\hat{\mathrm{k}})-(\hat{\mathrm{i}}+\hat{\mathrm{j}})\) \(=\hat{\mathrm{k}}-\hat{\mathrm{j}}\) We know that, \(\cos \theta=\frac{\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{AB}}}{|\overrightarrow{\mathrm{AC}}| \cdot|\overrightarrow{\mathrm{AB}}|}=\frac{(\hat{\mathrm{k}}-\hat{\mathrm{j}}) \cdot(\hat{\mathrm{k}}-\hat{\mathrm{i}})}{|(\hat{\mathrm{k}}-\hat{\mathrm{j}})||(\hat{\mathrm{k}}-\hat{\mathrm{i}})|}\) \(=\frac{\hat{\mathrm{k}} \hat{\mathrm{k}}-\hat{\mathrm{k}} \cdot \hat{\mathrm{i}}-\hat{\mathrm{j}} \hat{\mathrm{k}}+\hat{\mathrm{j}} \cdot \hat{\mathrm{i}}}{\sqrt{(1)^2+(-1)^2} \sqrt{(1)^2+(-1)^2}} \quad \because(\hat{\mathrm{k}} \cdot \hat{\mathrm{k}}=1)\) \(\cos \theta=\frac{1-0-0+0}{\sqrt{2} \sqrt{2}}=\frac{1}{2}\) \(\cos \theta=60^{\circ} \text { or } \theta=\frac{\pi}{3}\)
BITSAT-2018
Vector Algebra
87992
Let \(a, b\) and \(c\) be three vectors satisfying \(a \times b\) \(=(a \times c),|a|=|c|=1,|b|=4\) and \(|b \times c|=\sqrt{15}\). If \(b-2 c=\lambda a\), then \(\lambda\) equals
1 1
2 -1
3 2
4 -4
Explanation:
(D)Given, \(\mathrm{a} \times \mathrm{b}=(\mathrm{a} \times \mathrm{c}),\) \(|\mathrm{a}|=|\mathrm{c}|=1,\) \(|\mathrm{~b}|=4\) \(|\mathrm{~b} \times \mathrm{c}|=\sqrt{15}\) \(\mathrm{~b}-2 \mathrm{c}=\lambda \mathrm{a}\) \(\lambda=?\) Let \(\theta\) be the angle between \(b\) and \(c\). \(\because \quad|\mathrm{b} \times \mathrm{c}|=\sqrt{15}\) \(|\mathrm{~b}||\mathrm{c}| \sin \theta=\sqrt{15}\) \(\sin \theta=\frac{\sqrt{15}}{|\mathrm{~b}||\mathrm{c}|}\) \(\sin \theta=\frac{\sqrt{15}}{4 \times 1}=\frac{\sqrt{15}}{4}\) \(\therefore \cos \theta=\sqrt{1-\frac{15}{16}}=\frac{1}{\sqrt{16}}=\frac{1}{4}\) \(\mathrm{~b}-2 \mathrm{c}=\lambda \mathrm{a}\) (Given) Squaring in both sides, \(|\mathrm{b}-2 \mathrm{c}|^2=|\lambda \mathrm{a}|^2\) \(|\mathrm{~b}|^2+4|\mathrm{c}|^2-4(\mathrm{~b} \cdot \mathrm{c})=\lambda^2|\mathrm{a}|^2\) \(16+4-4|\mathrm{~b}||\mathrm{c}| \cos \theta=\lambda^2\) \(20-16 \cos \theta=\lambda^2\) \(20-16 \times \frac{1}{4}=\lambda^2\) \(20-4=\lambda^2\) \(\lambda^2=16\) \(\lambda= \pm 4\)
BITSAT-2015
Vector Algebra
87993
If \(|\vec{a}|=3,|\vec{b}|=4\), then a value of \(\lambda\), for which \(\vec{a}+\lambda \vec{b}\) is perpendicular to \(\vec{a}-\lambda \vec{b}\)
87991
If \(\hat{i}+\hat{j}, \hat{j}+\hat{k}, \hat{i}+\hat{k}\) are the position vectors of the vertices of a triangle \(A B C\) taken in order, then \(\angle \mathrm{A}\) is equal to
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{5}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{3}\)
Explanation:
(D) : Given position vector of the vertices, \(\overrightarrow{\mathrm{OA}}=\hat{i}+\hat{j} \Rightarrow \overrightarrow{\mathrm{OB}}=\hat{j}+\hat{k} \Rightarrow \overrightarrow{\mathrm{OC}}=\hat{i}+\hat{k}\) \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(\hat{j}+\hat{k})-(\hat{i}+\hat{j})\) \(=\hat{k}-\hat{i}\) \(\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=(\hat{\mathrm{i}}+\hat{\mathrm{k}})-(\hat{\mathrm{i}}+\hat{\mathrm{j}})\) \(=\hat{\mathrm{k}}-\hat{\mathrm{j}}\) We know that, \(\cos \theta=\frac{\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{AB}}}{|\overrightarrow{\mathrm{AC}}| \cdot|\overrightarrow{\mathrm{AB}}|}=\frac{(\hat{\mathrm{k}}-\hat{\mathrm{j}}) \cdot(\hat{\mathrm{k}}-\hat{\mathrm{i}})}{|(\hat{\mathrm{k}}-\hat{\mathrm{j}})||(\hat{\mathrm{k}}-\hat{\mathrm{i}})|}\) \(=\frac{\hat{\mathrm{k}} \hat{\mathrm{k}}-\hat{\mathrm{k}} \cdot \hat{\mathrm{i}}-\hat{\mathrm{j}} \hat{\mathrm{k}}+\hat{\mathrm{j}} \cdot \hat{\mathrm{i}}}{\sqrt{(1)^2+(-1)^2} \sqrt{(1)^2+(-1)^2}} \quad \because(\hat{\mathrm{k}} \cdot \hat{\mathrm{k}}=1)\) \(\cos \theta=\frac{1-0-0+0}{\sqrt{2} \sqrt{2}}=\frac{1}{2}\) \(\cos \theta=60^{\circ} \text { or } \theta=\frac{\pi}{3}\)
BITSAT-2018
Vector Algebra
87992
Let \(a, b\) and \(c\) be three vectors satisfying \(a \times b\) \(=(a \times c),|a|=|c|=1,|b|=4\) and \(|b \times c|=\sqrt{15}\). If \(b-2 c=\lambda a\), then \(\lambda\) equals
1 1
2 -1
3 2
4 -4
Explanation:
(D)Given, \(\mathrm{a} \times \mathrm{b}=(\mathrm{a} \times \mathrm{c}),\) \(|\mathrm{a}|=|\mathrm{c}|=1,\) \(|\mathrm{~b}|=4\) \(|\mathrm{~b} \times \mathrm{c}|=\sqrt{15}\) \(\mathrm{~b}-2 \mathrm{c}=\lambda \mathrm{a}\) \(\lambda=?\) Let \(\theta\) be the angle between \(b\) and \(c\). \(\because \quad|\mathrm{b} \times \mathrm{c}|=\sqrt{15}\) \(|\mathrm{~b}||\mathrm{c}| \sin \theta=\sqrt{15}\) \(\sin \theta=\frac{\sqrt{15}}{|\mathrm{~b}||\mathrm{c}|}\) \(\sin \theta=\frac{\sqrt{15}}{4 \times 1}=\frac{\sqrt{15}}{4}\) \(\therefore \cos \theta=\sqrt{1-\frac{15}{16}}=\frac{1}{\sqrt{16}}=\frac{1}{4}\) \(\mathrm{~b}-2 \mathrm{c}=\lambda \mathrm{a}\) (Given) Squaring in both sides, \(|\mathrm{b}-2 \mathrm{c}|^2=|\lambda \mathrm{a}|^2\) \(|\mathrm{~b}|^2+4|\mathrm{c}|^2-4(\mathrm{~b} \cdot \mathrm{c})=\lambda^2|\mathrm{a}|^2\) \(16+4-4|\mathrm{~b}||\mathrm{c}| \cos \theta=\lambda^2\) \(20-16 \cos \theta=\lambda^2\) \(20-16 \times \frac{1}{4}=\lambda^2\) \(20-4=\lambda^2\) \(\lambda^2=16\) \(\lambda= \pm 4\)
BITSAT-2015
Vector Algebra
87993
If \(|\vec{a}|=3,|\vec{b}|=4\), then a value of \(\lambda\), for which \(\vec{a}+\lambda \vec{b}\) is perpendicular to \(\vec{a}-\lambda \vec{b}\)
