jee-main 2019 Q87

jee-main · India · session2_09apr_shift1 Vectors: Cross Product & Distances

Let $\vec { \alpha } = 3 \hat { i } + \hat { j }$ and $\vec { \beta } = 2 \hat { i } - \hat { j } + 3 \hat { k }$. If $\vec { \beta } = \overrightarrow { \beta _ { 1 } } - \overrightarrow { \beta _ { 2 } }$, where $\overrightarrow { \beta _ { 1 } }$ is parallel to $\vec { \alpha }$ and $\overrightarrow { \beta _ { 2 } }$ is perpendicular to $\vec { \alpha }$, then $\overrightarrow { \beta _ { 1 } } \times \overrightarrow { \beta _ { 2 } }$ is equal to:
(1) $\frac { 1 } { 2 } ( - 3 \hat { i } + 9 \hat { j } + 5 \widehat { k } )$
(2) $3 \hat { i } - 9 \hat { j } - 5 \widehat { k }$
(3) $- 3 \hat { i } + 9 \hat { j } + 5 \widehat { k }$
(4) $\frac { 1 } { 2 } ( 3 \hat { i } - 9 \hat { j } + 5 \hat { k } )$

Let $\vec { \alpha } = 3 \hat { i } + \hat { j }$ and $\vec { \beta } = 2 \hat { i } - \hat { j } + 3 \hat { k }$. If $\vec { \beta } = \overrightarrow { \beta _ { 1 } } - \overrightarrow { \beta _ { 2 } }$, where $\overrightarrow { \beta _ { 1 } }$ is parallel to $\vec { \alpha }$ and $\overrightarrow { \beta _ { 2 } }$ is perpendicular to $\vec { \alpha }$, then $\overrightarrow { \beta _ { 1 } } \times \overrightarrow { \beta _ { 2 } }$ is equal to:\\
(1) $\frac { 1 } { 2 } ( - 3 \hat { i } + 9 \hat { j } + 5 \widehat { k } )$\\
(2) $3 \hat { i } - 9 \hat { j } - 5 \widehat { k }$\\
(3) $- 3 \hat { i } + 9 \hat { j } + 5 \widehat { k }$\\
(4) $\frac { 1 } { 2 } ( 3 \hat { i } - 9 \hat { j } + 5 \hat { k } )$

Paper Questions

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